State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 1

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 2

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 3

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 4

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 5

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 6

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 7

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 8

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 9

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 10

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 11

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 12

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 13

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 14

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 15

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 16

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 17

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 18

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 19

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 20

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 21

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 22

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 23

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 24

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 25

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 26

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 27

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 28

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 29

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 30

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 31

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 32

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 33

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 34

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 35

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 36

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

eSolutions Manual - Powered by Cognero Page 37

Study Guide and Review - Chapter 8

State whether each sentence is true or false . If false , replace the underlined phrase or expression to make a true sentence.

1. x2 + 5x + 6 is an example of a prime polynomial.

SOLUTION:  The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial.

2. (x + 5)(x − 5) is the factorization of a difference of squares.

SOLUTION:  

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is the factorization of a difference of squares. The statement is true.

3. (x + 5)(x − 2) is the factored form of x2 − 3x − 10.

SOLUTION:  

(x − 5)(x + 2) is the factored form of x2 − 3x − 10. So, the statement is false.

4. Expressions with four or more unlike terms can sometimes be factored by grouping.

SOLUTION:  Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping becauseterms are put into groups and then factored. So, the statement is true.

5. The Zero Product Property states that if ab = 1, then a or b is 1.

SOLUTION:  The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b = 0, or a and b are zero.

6. x2 − 12x + 36 is an example of a perfect square trinomial.

SOLUTION:  

Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

7. x2 − 16 is an example of a perfect square trinomial.

SOLUTION:  

The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.

x2 − 16 is the product of a sum and a difference. So, x

2 − 16 an example of a difference of squares.

8. 4x2 – 2x + 7 is a polynomial of degree 2.

SOLUTION:  true

9. The leading coefficient of 1 + 6a + 9a2 is 1.

SOLUTION:  The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

10. The FOIL method is used to multiply two trinomials.

SOLUTION:  false; binomials   FOIL Method:   To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L and the Last terms.

Write each polynomial in standard form.

11. x + 2 + 3x2

SOLUTION:  

The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x

2 + x + 2.

12. 1 – x4

SOLUTION:  

The greatest degree is 4. Therefore, the polynomial can be rewritten as –x

4 + 1.

13. 2 + 3x + x2

SOLUTION:  

 

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

14. 3x5 – 2 + 6x – 2x

2 + x

3

SOLUTION:  

 

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x

3 – 2x

2 + 6x – 2.

Find each sum or difference.

15. (x3 + 2) + (−3x

3 − 5)

SOLUTION:  

16. a2 + 5a − 3 − (2a2 − 4a + 3)

SOLUTION:  

17. (4x − 3x2 + 5) + (2x

2 − 5x + 1)

SOLUTION:  

18. PICTURE FRAMES  Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

SOLUTION:  

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

19. x2(x + 2) = x(x

2 + 2x + 1)

SOLUTION:  

 

20. 2x(x + 3) = 2(x2 + 3)

SOLUTION:  

21. 2(4w + w2) − 6 = 2w(w − 4) + 10

SOLUTION:  

22. GEOMETRY  Find the area of the rectangle.

SOLUTION:  

The area of the rectangle is 3x3 + 3x

2 − 21x.

Find each product.

23. (x − 3)(x + 7)

SOLUTION:  

24. (3a − 2)(6a + 5)

SOLUTION:  

25. (3r − 7t)(2r + 5t)

SOLUTION:  

26. (2x + 5)(5x + 2)

SOLUTION:  

27. PARKING LOT  The parking lot shown is to be paved. What is the area to be paved?

SOLUTION:  

The area to be paved is 10x2 + 7x − 12 units

2.

Find each product.

28. (x + 5)(x − 5)

SOLUTION:  

29. (3x − 2)2

SOLUTION:  

30. (5x + 4)2

SOLUTION:  

31. (2x − 3)(2x + 3)

SOLUTION:  

32. (2r + 5t)2

SOLUTION:  

33. (3m − 2)(3m + 2)

SOLUTION:  

34. GEOMETRY  Write an expression to represent the area of the shaded region.

SOLUTION:  Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger rectangle.

The area of the shaded region is 3x2 − 21.

Use the Distributive Property to factor each polynomial.35. 12x + 24y

SOLUTION:  Factor

The greatest common factor of each term is 2· 3 or 12.12x + 24y = 12(x + 2y)

36. 14x2y − 21xy + 35xy

2

SOLUTION:  Factor

The greatest common factor of each term is 7xy.

14x2y − 21xy + 35xy

2 = 7xy(2x − 3 + 5y)

37. 8xy − 16x3y + 10y

SOLUTION:  Factor.

The greatest common factor of each term is 2y .

8xy − 16x3y + 10y = 2y(4x − 8x

3 + 5)

38. a2 − 4ac + ab − 4bc

SOLUTION:  Factor by grouping.

39. 2x2 − 3xz − 2xy + 3yz

SOLUTION:  

40. 24am − 9an + 40bm − 15bn

SOLUTION:  

Solve each equation. Check your solutions.

41. 6x2 = 12x

SOLUTION:  Factor the trinomial using the Zero Product Property.

       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

42. x2 = 3x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.

and

The solutions are 0 and 3.

43. 3x2 = 5x

SOLUTION:  Factor the trinomial using the Zero Product Property.

The roots are 0 and . Check by substituting 0 and in for x in the original equation.

and

The solutions are 0 and .

44. x(3x − 6) = 0

SOLUTION:  Factor the trinomial using the Zero Product Property. x(3x − 6) = 0

x = 0       or       

The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation.

and

The solutions are 0 and 2.

45. GEOMETRY  The area of the rectangle shown is x3 − 2x

2 + 5x square units. What is the length?

SOLUTION:  

The area of the rectangle is x3 − 2x

2 + 5x or x(x

2 – 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 − 2x + 5.

Factor each trinomial. Confirm your answers using a graphing calculator.

46. x2 − 8x + 15

SOLUTION:  

In this trinomial, b = –8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be negative. List the negative factors of 15, and look for the pair of factors with a sum of –8.  

  The correct factors are –3 and –5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 15 Sum of –8 –1, –15 –16 –3, –5 –8

47. x2 + 9x + 20

SOLUTION:  

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive. List the positive factors of 20, and look for the pair of factors with a sum of 9.  

  The correct factors are 4 and 5.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–10, 10] scl: 1

Factors of 20 Sum of 9 1, 20 21 2, 10 12 4, 5 9

48. x2 − 5x − 6

SOLUTION:  

In this trinomial, b = –5 and c = –6, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of –6, and look for the pair of factors with a sum of –5.  

  The correct factors are 1 and −6.

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–12, 8] scl: 1

Factors of –6 Sum of –5 –1, 6 5 1, –6 –5 2, –3 –1 –2, 3 1

49. x2 + 3x − 18

SOLUTION:  

In this trinomial, b = 3 and c = –18, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of –18, and look for the pair of factors with a sum of 3.  

  The correct factors are –3 and 6.  

  Check using a Graphing calculator.

[–10, 10] scl: 1 by [–14, 6] scl: 1

Factors of –18 Sum of 3 –1, 18 17 1, –18 –17 –2, 9 7 2, –9 –7 –3, 6 3 3, –6 –3

Solve each equation. Check your solutions.

50. x2 + 5x − 50 = 0

SOLUTION:  

The roots are –10 and 5. Check by substituting –10 and 5 in for x in the original equation.

and

The solutions are –10 and 5.

51. x2 − 6x + 8 = 0

SOLUTION:  

The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation.

and

The solutions are 2 and 4.

52. x2 + 12x + 32 = 0

SOLUTION:  

The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.

and

The solutions are –8 and –4.

53. x2 − 2x − 48 = 0

SOLUTION:  

The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.

and

The solutions are –6 and 8.

54. x2 + 11x + 10 = 0

SOLUTION:  

The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.

and

The solutions are –10 and –1.

55. ART  An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square inches. What is the length of the painting?

SOLUTION:  Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime .

56. 12x2 + 22x − 14

SOLUTION:  

In this trinomial, a = 12, b = 22 and c = –14, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 12(–14) or –168 and identify the factors with a sum of 22.

The correct factors are –6 and 28.

So, 12x2 + 22x − 14 = 2(2x − 1)(3x + 7).

Factors of –168 Sum  –2, 84 82 2, –84 –82 –3, 56 53 3, –56 –53 –4, 42 38 4, –42 –38 –6, 28 22

57. 2y2 − 9y + 3

SOLUTION:  

In this trinomial, a = 2, b = –9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be negative.   2(3) = 6   There are no factors of 6 with a sum of –9. So, this trinomial is prime.

58. 3x2 − 6x − 45

SOLUTION:  

In this trinomial, a = 3, b = –6 and c = –45, so m + p is negative and mp is negative. Therefore, m and p must have different signs. List the factors of 3(–45) or –135 and identify the factors with a sum of –6.

The correct factors are –15 and 9.

So, 3x2 − 6x − 45 = 3(x − 5)(x + 3).

Factors of –135 Sum –1, 135 134 1, –135 –134 –3, 45 42 3, –45 –42 –5, 27 22 5, –27 –22 –9, 15 6 9, –15 –6

59. 2a2 + 13a − 24

SOLUTION:  

In this trinomial, a = 2, b = 13 and c = –24, so m + p is positive and mp is negative. Therefore, m and p must have different signs. List the factors of 2(–24) or –48 and identify the factors with a sum of 13.

The correct factors are –3 and 16.

So, 2a2 + 13a − 24 = (2a − 3)(a + 8).

Factors of –48 Sum  –1, 48 47 1, –48 –47 –2, 24 22 2, –24 –22 –3, 16 13 3, –16 –13 –4, 12 8 4, –12 –8 –6, 8 2 6, –8 –2

Solve each equation. Confirm your answers using a graphing calculator.

60. 40x2 + 2x = 24

SOLUTION:  

The roots are  and  or –0.80 and 0.75 . 

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = –24. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 1 by [–5, 25] scl: 3

[–5, 5] scl: 1 by [–5, 25] scl: 3

61. 2x2 − 3x − 20 = 0

SOLUTION:  

The roots are or −2.5 and 4. 

Confirm the roots using a graphing calculator. Let Y1 = 2x2 − 3x − 20 and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are  and 4.

  [–10, 10] scl: 1 by [–15, 5] scl: 1

[–10, 10] scl: 1 by [–15, 5] scl: 1

62. −16t2 + 36t − 8 = 0

SOLUTION:  

The roots are 2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = −16t2 + 36t − 8  and Y2 = 0. Use the intersect option from

the CALC menu to find the points of intersection. 

The solutions are 2 and .

  [–2, 3] scl: 1 by [–20, 10] scl: 6

[–2, 3] scl: 1 by [–20, 10] scl: 6

63. 6x2 − 7x − 5 = 0

SOLUTION:  

The roots are  and  or −0.5 and 1.67. 

 

Confirm the roots using a graphing calculator. Let Y1 = 6x2 − 7x − 5 and Y2 = 0. Use the intersect option from the

CALC menu to find the points of intersection. 

The solutions are  and  .

  [–5, 5] scl: 0.5 by [–10, 10] scl: 1

[–5, 5] scl: 0.5 by [–10, 10] scl: 1

64. GEOMETRY  The area of the rectangle shown is 6x2 + 11x − 7 square units. What is the width of the rectangle?

SOLUTION:  To find the width, factor the area of the rectangle. In the area trinomial, a = 6, b = 11 and c = –7, so m + p is positive and mp is negative. Therefore, m and p must havedifferent signs. List the factors of 6(–7) or –42 and identify the factors with a sum of 11.

The correct factors are –3 and 14.

So, 6x2 + 11x − 7 = (2x – 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x – 1, the width must be 3x + 7.

Factors of –42 Sum –1, 42 41 1, –42 –41 –2, 21 19 2, –21 –19 –3, 14 11 3, –14 –11 –6, 7 1 6, –7 –1

Factor each polynomial.

65. y2 − 81

SOLUTION:  

66. 64 − 25x2

SOLUTION:  

67. 16a2 − 21b

2

SOLUTION:  

The number 21 is not a perfect square. So, 16a2 − 21b

2 is prime.

68. 3x2 − 3

SOLUTION:  

Solve each equation by factoring. Confirm your answers using a graphing calculator.

69. a2 − 25 = 0

SOLUTION:  

The roots are –5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

      Thus, the solutions are –5 and 5.

70. 9x2 − 25 = 0

SOLUTION:  

   

 

The roots are  and  or about -1.667 and 1.667. 

  Confirm the roots using a graphing calculator. Let Y1 = 9x2 – 25 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.  

        

Thus, the solutions are  and  .

71. 81 − y2 = 0

SOLUTION:   

   

  The roots are -9 and 9.    Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the CALC menu to find the points of intersection.

          [-10, 10] scl: 1 by [-10, 90] scl:10                [-10, 10] scl: 1 by [-10, 90] scl:10   Thus, the solutions are –9 and 9.

72. x2 − 5 = 20

SOLUTION:  

   

The roots are -5 and 5.    Confirm the roots using a graphing calculator. Let Y1 = x2 - 5 and Y2 = 20. Use the intersect option from the CALC menu to find the points of intersection.\  

         [-10, 10] scl:1 by [-10, 40] scl: 5                    [-10, 10] scl:1 by [-10, 40] scl: 5   Thus, the solutions are –5 and 5.

73. EROSION  A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t

seconds is given by the equation d = 16t2. How long does it take the boulder to hit the water?

SOLUTION:  

The distance the boulder falls is 64 feet. So, 64 = 16t2.

 

   

  The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.

Factor each polynomial, if possible. If the polynomial cannot be factored write prime .

74. x2 + 12x + 36

SOLUTION:  

75. x2 + 5x + 25

SOLUTION:  

There are no factors of 25 that have a sum of 5. So, x2 + 5x + 25 is prime.

76. 9y2 − 12y + 4

SOLUTION:  

77. 4 − 28a + 49a2

SOLUTION:  

78. x4 − 1

SOLUTION:  

79. x4 − 16x

2

SOLUTION:  

Solve each equation. Confirm your answers using a graphing calculator.

80. (x − 5)2 = 121

SOLUTION:  

   

  The roots are –6 and 16.   

Confirm the roots using a graphing calculator. Let Y1 = (x − 5)2 and Y2 = 121. Use the intersect option from the

CALC menu to find the points of intersection.   

  The solutions are –6 and 16.

  [–20, 20] scl: 3 by [–5, 125] scl: 13

[–20, 20] scl: 3 by [–5, 125] scl: 13

81. 4c2 + 4c + 1 = 9

SOLUTION:  

   

  The roots are –2 and 1.   

Confirm the roots using a graphing calculator. Let Y1 = 4c2 + 4c + 1 and Y2 = 9. Use the intersect option from the

CALC menu to find the points of intersection.   

  Thus, the solutions are –2 and 1.

  [–5, 5] scl: 1 by [–5, 15] scl: 2

[–5, 5] scl: 1 by [–5, 15] scl: 2

82. 4y2 = 64

SOLUTION:  

   

  The roots are –4 and 4.   

Confirm the roots using a graphing calculator. Let Y1 = 4y2 and Y2 = 64. Use the intersect option from the CALC

menu to find the points of intersection.   

  Thus, the solutions are –4 and 4.

  [–10, 10] scl: 1 by [–5, 75] scl: 10

[–10, 10] scl: 1 by [–5, 75] scl: 10

83. 16d2 + 40d + 25 = 9

SOLUTION:  

   

 

The roots are –2 and . 

 

Confirm the roots using a graphing calculator. Let Y1 = 16d2 + 40d + 25 and Y2 = 9. Use the intersect option from

the CALC menu to find the points of intersection.   

 

Thus, the solutions are –2 and .

  [–5, 5] scl: 0.5 by [–5, 15] scl: 5

[–5, 5] scl: 0.5 by [–5, 15] scl: 5

84. LANDSCAPING  A sidewalk is being built around a square yard that is 25 feet on each side. The total area of the yard and sidewalk is 900 square feet. What is the width of the sidewalk?

SOLUTION:  Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the width and length are the same.  

 

 

  The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.

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Study Guide and Review - Chapter 8