polynomial functions

ALGEBRA 2 LESSON 6-1ALGEBRA 2 LESSON 6-1

(For help, go to Lesson 1-2.)

Polynomial FunctionsPolynomial Functions

Simplify each expression by combining like terms.

1. 3x + 5x – 7x 2. –8xy2 – 2x2y + 5x2y 3. –4x + 7x2 + x

Find the number of terms in each expression.

4. bh 5. 1 – x 6. 4x3 – x2 – 912

6-1

1. 3x + 5x – 7x = (3 + 5 – 7)x = 1x = x

2. –8xy2 – 2x2y + 5x2y = (–2 + 5)x2y – 8xy2 = 3x2y – 8xy2

3. –4x + 7x2 + x = 7x2 + (–4 + 1)x = 7x2 – 3x

4. bh; 1 term

5. 1 – x; 2 terms

6. 4x3 – x2 – 9; 3 terms

12



Solutions

6-1



Write each polynomial in standard form. Then classify it by degree and by number of terms.

a. 9 + x3 b. x3 – 2x2 – 3x4

x3 + 9 –3x4 + x3 – 2x2

The polynomial is a quartic trinomial.

The term with the largest degree is x3,so the polynomial is degree 3.

It has two terms.The polynomial is a cubic binomial.

The term with the largest degree is –3x4, so the polynomial is degree 4. It has three terms.

6-1

x y0 2.82 54 66 5.58 4

Using a graphing calculator, determine whether a linear,

quadratic, or cubic model best fits the values in the table.



Enter the data. Use the LinReg, QuadReg, and CubicReg options of a graphing calculator to find the best-fitting model for each polynomial classification.

Graph each model and compare.

The quadratic model appears to best fit the given values.

Linear model Quadratic model Cubic model

6-1

To estimate the number of employees for 1988, you can use the Table function option of a graphing calculator to find that ƒ(13) 62.72. According to the model, there were about 62 employees in 1988.

The table shows data on the number of employees that a small

company had from 1975 to 2000. Find a cubic function to model the data.

Use it to estimate the number of employees in 1998. Let 0 represent 1975.



To find a cubic model, use the CubicReg option of a graphing calculator.

The function ƒ(x) = 0.0096x3 – 0.375x2 + 3.541x + 58.96 is an approximate model for the cubic function.

1975 60

1980 65

1985 70

1990 60

1995 55

2000 64

Number ofEmployeesYear

Enter the data.Graph the model.

6-1



pages 303–305 Exercises

1. 10x + 5; linear binomial

2. –3x + 5; linear binomial

3. 2m2 + 7m – 3; quadratic binomial

4. x4 – x3 + x; quartic trinomial

5. 2p2 – p; quadratic binomial

6. 3a3 + 5a2 + 1; cubic trinomial

7. –x5; quintic monomial

8. 12x4 + 3; quartic binomial

9. 5x3; cubic monomial

10. –2x3; cubic monomial

11. 5x2 + 4x + 8; quadratic trinomial

12. –x4 + 3x3; quartic binomial

13. y = x3 + 1

14. y = 2x3 – 12

15. y = 1.5x3 + x2 – 2x + 1

16. y = –3x3 – 10x2 + 100

17. a. males: y = –0.002571x2 + 0.2829x + 67.21females: y = –0.002286x2 + 0.2514x + 74.82

b. males: y = 0.00008333x3 – 0.007571x2 + 0.3545x + 67.11females: y = 0.00008333x3 – 0.007286x2 + 0.3231x + 74.72

c. The cubic model is a better fit.

6-1



18. y = x3 – 2x2; 4335

19. y = x3 – 10x2; 2023

20. y = –0.5x3 + 10x2; 433.5

21. y = –0.03948x3 + 2.069x2 – 17.93x + 106.9; 206.07

22. y = –0.007990x3 + 0.4297x2 – 6.009x + 43.57; 26.34

23. y = 0.01002x3 – 0.3841x2 + 5.002x + 2.132; 25.39

24. Check students’ work.

25. x3 + 4x; cubic binomial

26. –4a4 + a3 + a2; quartic trinomial

27. 7; constant monomial

28. 6x2; quadratic monomial

29. x4 + 2x3; quartic binomial

30. x5 + x; quintic binomial

31. a. V = 10 r 2

b. V = r 3

c. V = r 3 + 10 r 2

32. Answers may vary. Sample: Cubic functions represent curvature in the data. Because of their turning points they can be unreliable for extrapolation.

33. –c2 + 16; binomial

34. –9d3 – 13; binomial

23

12

2323

6-1



35. 16x2 – x – 5; trinomial

36. 2x3 – 6x + 17; trinomial

37. a + 4b; binomial

38. –12y; monomial

39. 8x2 – 6y; binomial

40. –3a + 2; binomial

41. 2x3 + 9x2 + 5x + 27; polynomial of 4 terms

42. –4x4 – 3x3 + 5x – 54; polynomial of 4 terms

43. 80x3 – 109x2 + 7x – 75; polynomial of 4 terms

44. 2x3 – 2x2 + 8x – 27; polynomial of 4 terms

45. 6a2 + 3ab – 8; trinomial

46. 8x3 + 2x2; binomial

47. 30x3 – 10x2; binomial

48. 2a3 – 5a2 – 2a + 5; polynomial of 4 terms

49. b3 – 6b2 + 9b; trinomial

50. x3 – 6x2 + 12x – 8; polynomial of 4 terms

51. x4 + 2x2 + 1; trinomial

52. 8x3 + 60x2 + 150x + 126; polynomial of 4 terms

53. a3 – a2b – b2a + b3; polynomial of 4 terms

6-1



54. a4 – 4a3 + 6a2 – 4a + 1; polynomial of 5 terms

55. 12s3 + 61s2 + 68s – 21; polynomial of 4 terms

56. x3 + 2x2 – x – 2; trinomial

57. 8c3 – 26c + 12; trinomial

58. s4 – 2t 2s2 + t 4; trinomial

59. a. y = 0.7166x + 47.61y = 0.0009365x3 – 0.0744x2 +

2.2929x + 41.4129y = –0.00004789x4 + 0.004666x3 –

0.1647x2 + 2.9797x + 40.7831

b.

The quartic model fits best.

c. 72.2 1015

6-1



60. 2.5 108 cm3

61. a. up 4 unitsb. y = 4x3 is more narrow.c. y = x3

62. B

63. A

64. A

65. [2] If it is in standard form, the degree is the exponent on the first variable.

[1] incorrect reason for why it is easier to find in standard form

66. 2

67. 2

68. none

69.

(5, –3)

70. 1 –3 1 5–8 –3 0 –3

71. 2 –3 –65 –2 0

72. –2 –1 –2 –1–3 –3 –4 –4

6-1

Write each polynomial in standard form. Then classify it by degree and by number of terms.

1. –x2 + 2x + x2 2. 7x2 + 10 + 4x3 3. 3x(4x) + x2(2x2)

4.

a. Find a cubic and a quadratic model for the table of values.

b. Which model appears to give the better fit?

c. Using the model you selected in part b, estimate the value of y when x = 12.



x 1 2 3 5 8

y 3 2.2 3 7 5

2x; degree 1, linear monomial

2x4 + 12x2; degree 4, quartic binomial

4x3 + 7x2 + 10; degree 3, cubic trinomial

the cubic model

–49.3

y = –0.14401x3 + 1.7805x2 – 5.29x + 6.62,y = –0.1319x2 + 1.682x + 0.37

6-1

Factor each quadratic expression.

1. x2 + 7x + 12 2. x2 + 8x – 20 3. x2 – 14x + 24

Find each product.

4. x(x + 4) 5. (x + 1)2 6. (x – 3)2(x + 2)

(For help, go to Lessons 5-1 and 5-4.)


Polynomials and Linear FactorsPolynomials and Linear Factors

6-2

Solutions



1. Factors of 12 with a sum of 7: 4 and 3x2 + 7x + 12 = (x + 4)(x + 3)

2. Factors of –20 with a sum of 8: 10 and –2x2 + 8x – 20 = (x + 10)(x – 2)

3. Factors of 24 with a sum of –14: –12 and –2x2 – 14x + 24 = (x – 12)(x – 2)

4. x(x + 4) = x(x) + x(4) = x2 + 4x

5. (x + 1)2 = x2 + 2(1)x + 12 = x2 + 2x + 1

6. (x – 3)2(x + 2) = (x2 – 2(3)x + 32)(x + 2)= (x2 – 6x + 9)(x + 2)= (x2 – 6x + 9)(x) + (x2 – 6x + 9)(2)= (x3 – 6x2 + 9x) + (2x2 – 12x + 18)= x3 + (–6 + 2)x2 + (9 – 12)x + 18= x3 – 4x2 – 3x + 18

6-2

Write (x – 1)(x + 3)(x + 4) as a polynomial in standard form.



(x – 1)(x + 3)(x + 4) = (x – 1)(x2 + 4x + 3x + 12) Multiply (x + 3) and (x + 4).

= (x – 1)(x2 + 7x + 12) Simplify.

= x(x2 + 7x + 12) – 1(x2 + 7x + 12) Distributive Property

= x3 + 7x2 + 12x – x2 – 7x – 12 Multiply.

= x3 + 6x2 + 5x – 12 Simplify.

The expression (x – 1)(x + 3)(x + 4) is the factored form of x3 + 6x2 + 5x – 12.

6-2

Write 3x3 – 18x2 + 24x in factored form.



3x3 – 18x2 + 24x = 3x(x2 – 6x + 8) Factor out the GCF, 3x.

= 3x(x – 4)(x – 2) Factor x2 – 6x + 8.

Check: 3x(x – 4)(x – 2) = 3x(x2 – 6x + 8) Multiply (x – 4)(x – 2).

= 3x3 – 18x2 + 24x Distributive Property

6-2

Another airline has different carry-on luggage regulations. The sum of the length, width, and depth may not exceed 50 in.



a. Assume that the sum of the length, width, and depth is 50 in. and the length is 10 in. greater than the depth. Graph the function relating the volume v to depth x. Find the x-intercepts. What do they represent?

Relate: Volume = depth • length • width

Define: Let x = depth. Then x + 10 = length, and

50 – (depth + length) = width.

Write: V(x) = x ( x + 10 )( 50 – (x + x + 10) )

= x (x + 10)(40 – 2x)

The x-intercepts of the function are x = 0, x = –10, x = 20. These values of x produce a volume of zero.

Graph the function for volume.

6-2

(continued)



b. Describe a realistic domain for V(x).

c. What is the maximum possible volume of the box? What are the corresponding dimensions of the box?

Since the volume must be positive, x < 20. A realistic domain is 0 < x < 20.

Then the length is about 22.2 in. and the width is about 15.6 in.

The function has values over the set of all real numbers x.Since x represents the depth of the luggage, x > 0.

Look for the greatest value of y that occurs within the domain 0 < x < 20.

Use the Maximum feature of a graphing calculator to find the maximum volume.

A volume of approximately 4225 in.3 occurs for a depth of about 12.2 in.

6-2

Find the zeros of y = (x + 1)(x – 1)(x + 3). Then graph the

function using a graphing calculator.



Using the Zero Product Property, find a zero for each linear factor.

x + 1 = 0 or x – 1 = 0 or x + 3 = 0x = –1 x = 1 x = –3

The zeros of the function are –1, 1, –3.

Now sketch and label the function.

6-2

Write a polynomial in standard form with zeros at 2, –3, and 0.



= (x – 2)(x2 + 3x) Multiply (x + 3)(x).

= x(x2 + 3x) – 2(x2 + 3x) Distributive Property

= x3 + 3x2 – 2x2 – 6x Multiply.

= x3 + x2 – 6x Simplify.

The function ƒ(x) = x3 + x2 – 6x has zeros at 2, –3, and 0.

2 –3 0 Zeros

ƒ(x) = (x – 2)(x + 3)(x) Write a linear factor for each zero.

6-2

Find any multiple zeros of ƒ(x) = x5 – 6x4 + 9x3 and state the

multiplicity.



ƒ(x) = x5 – 6x4 + 9x3

ƒ(x) = x3(x2 – 6x + 9) Factor out the GCF, x3.

ƒ(x) = x3(x – 3)(x – 3) Factor x2 – 6x + 9.

Since you can rewrite x3 as (x – 0)(x – 0)(x – 0), or (x – 0)3, the number 0 is a multiple zero of the function, with multiplicity 3.

Since you can rewrite (x – 3)(x – 3) as (x – 3)2, the number 3 is a multiple zero of the function with multiplicity 2.

6-2




1. x2 + x – 6

2. x3 + 12x2 + 47x + 60

3. x3 – 7x2 + 15x – 9

4. x3 + 4x2 + 4x

5. x3 + 10x2 + 25x

6. x3 – x

7. x(x – 6)(x + 6)

8. 3x(3x – 1)(x + 1)

9. 5x(2x2 – 2x + 3)

10. x(x + 5)(x + 2)

11. x(x + 4)2

12. x(x – 9)(x + 2)

13. 24.2, –1.4, 0, –5, 1

14. 5.0, –16.9, 2, 6, 8

15. a. h = x, = 16 – 2x, w = 12 – 2x

b. V = x(16 – 2x)(12 – 2x)

c.

194 in.3, 2.26 in.

16. 1, –2

17. 2, –9

18. 0, –5, 8

19. –1, 2, 3

6-2



20. –1, 1, 2

21. y = x3 – 18x2 + 107x – 210

22. y = x3 + x2 – 2x

23. y = x3 + 9x2 + 15x – 25

24. y = x3 – 9x2 + 27x – 27

25. y = x3 + 2x2 – x – 2

26. y = x3 + 6x2 + 11x + 6

27. y = x3 – 2x2

28. y = x3 – x2 – 2x

29. –3 (mult. 3)

30. 0, 1 (mult. 3)

31. –1, 0,

32. –1, 0, 1

33. 4 (mult. 2)

34. 1, 2 (mult. 2)

35. – , 1 (mult. 2)

36. –1 (mult. 2), 1, 2

37. 2 x3 blocks, 15 x2 blocks, 31 x blocks, 12 unit blocks

38. a. V = 2x3 + 15x2 + 31x + 12; 2x3 + 7x2 + 7x + 2

b. V = 8x2 + 24x + 10

39. V = 12x3 – 27x

12

32

6-2

72



40. a. h = x + 3; w = xb.

0 < –3 < 2; where the volume is zeroc. 0 < x < 2d. 4.06 ft3

41. y = –2x3 + 9x2 – x – 12

42. y = 5x4 – 23x3 – 250x2 + 1164x + 504

43. y = 3x(x – 8)(x – 1)

44. y = –2x(x + 5)(x – 4)

45. y = x2(x + 4)(x – 1)

46. y = x x – x +

47. 2.5, 5.1; , 4, 6

48. 0.9, –6.9, –1.4; 0, –3, –1, 1

49. 2.98, –6.17; 1.5

50. none, –1; –2, 0

51–53. Answers may vary. Samples are given.

51. y = x3 – 3x2 – 10x

52. y = x3 – 21x2 + 147x – 343

53. y = x4 – 4x3 – 7x2 + 22x + 24

54. –4, 5 (mult. 3)

55. 0 (mult. 2), –1 (mult. 2)

12

12

12

6-2

32



56. 0, 6, –6

57. Answers may vary. Sample: Write the polynomial in standard form. The constant term is the value of the y-intercept.

58. 1 ft

59. Answers may vary. Sample: y = x4 – x2, and zeros are 0, ±1.

60. Answers may vary. Sample: The linear factors can be determined by examining the x-intercepts of the graph.

61. x + 2a

62. a. A = –x3 + 2x2 + 4xb. 6 square units

63. Answers may vary. Sample: y = (x – 1)(x + 1)(x – i )(x + i ); y = x4 – 1

64. a. Answers may vary. Sample: translation to the right 4 units

b. No; the second graph is not the result of a horizontal translation.

c. Answers may vary. Sample: rotation of 180° about the origin

65. C

66. H

67. B

68. [2] ƒ(x) = x2(x + 9)(x – 1)[1] incomplete factoring of x2 + 8x – 9

or other minor errors

6-2

78



69. [4] The student multiplies (x + 2), (x – 5), (x – 6) and –2 to get y = –2x3 + 18x2 – 16x – 120.

[3] appropriate methods but with one computational error

[2] appropriate methods to find basic polynomial with given roots, but no multiplication of polynomial by –2

[1] correct function, without work shown

70. –3x5 + 3x2 – 1; quintic trinomial

71. –7x4 – x3; quartic binomial

72. x3 – 2; cubic binomial

73. (x + 4)(x + 1)

74. (x – 5)(x + 3)

75. (x – 6)2

76. 8

77. –11

78. 0

6-2



1. Find the zeros of y = (x + 7)(x – 3)(x – 2). Then write the polynomialin standard form.

2. Factor x3 – 2x2 – 15x. Find the relative maximum and relative minimum.

3. Write a polynomial function in standard form with zeros at –5, –4, and 3.

4. Find any multiple zeros of ƒ(x) = x4 – 25x2 and state the multiplicity.

–7, 2, 3; x3 + 2x2 – 29x + 42

Answers may vary. Sample: ƒ(x) = x3 + 6x2 – 7x – 60

The number 0 is a zero with multiplicity 2.

6-2

x(x + 3)(x – 5); , –3640027



Dividing PolynomialsDividing Polynomials

Simplify each expression.

1. (x + 3)(x – 4) + 2 2. (2x + 1)(x – 3)

3. (x + 2)(x + 1) – 11 4. –3(2 – x)(x + 5)

Write each polynomial in standard form. Then list the coefficients.

5. 5x – 2x2 + 9 + 4x3 6. 10 + 5x3 – 9x2

7. 3x + x2 – x + 7 – 2x2 8. –4x4 – 7x2 + x3 + x4

6-3

Solutions



1. (x + 3)(x – 4) + 2 = x2 – 4x + 3x – 12 + 2= x2 – x – 10

2. (2x + 1)(x – 3) = 2x2 – 6x + x – 3 = 2x2 – 5x – 3

3. (x + 2)(x + 1) – 11 = x2 + x + 2x + 2 – 11 = x2 + 3x – 9

4. –3(2 – x)(x + 5) = –3(2x + 10 – x2 – 5x) = –3(–x2 – 3x + 10) = 3x2 + 9x – 30

5. 5x – 2x2 + 9 + 4x3 = 4x3 – 2x2 + 5x + 9 coefficients: 4, –2, 5, 9

6. 10 + 5x3 – 9x2 = 5x3 – 9x2 + 10 coefficients: 5, –9, 0, 10

7. 3x + x2 – x + 7 – 2x2 = –x2 + 2x + 7 coefficients: –1, 2, 7

8. –4x4 – 7x2 + x3 + x4 = –3x4 + x3 – 7x2 coefficients: –3, 1, –7, 0, 0

6-3

Divide x2 + 2x – 30 by x – 5.



– 30 Subtract: (x2 + 2x) – (x2 – 5x) = 7x. Bring down –30.

x Divide = x.x – 5 x2 + 2x – 30

x2

x

x2 – 5x Multiply: x(x – 5) = x2 – 5x 7x

Repeat the process of dividing, multiplying, and subtracting.

5 Subtract: (7x – 30) – (7x – 35) = 5.

The quotient is x + 7 with a remainder of 5, or simply x + 7, R 5.

7x – 35 Multiply: 7(x – 5) = 7x – 35.

x + 7 Divide = 7.x – 5 x2 + 2x – 30

x2 – 5x7x – 30

7xx

6-3

Determine whether x + 2 is a factor of each polynomial.



a. x2 + 10x + 16 b. x3 + 7x2 – 5x – 6

Since the remainder is zero,x + 2 is a factor of x2 + 10x + 16.

x + 8x + 2 x2 + 10x + 16

x2 + 2x8x + 16 8x + 16

0

x2 + 5x – 15x + 2 x3 + 7x2 – 5x – 6

x3 + 2x2

5x2 – 5x5x2 + 10x

–15x – 6–15x – 30

24

Since the remainder = 0, x + 2 is not a factor of x3 + 7x2 – 5x – 6.

/

6-3

Use synthetic division to divide 5x3 – 6x2 + 4x – 1 by x – 3.



Step 1: Reverse the sign of the constant term in the divisor. Write the coefficients of the polynomial in standard form.

Step 2: Bring down the coefficient.

Write x – 3 5x3 – 6x2 + 4x – 1

as 3 5 –6 4 –1

Bring down the 5.3 5 –6 4 –1 This begins the quotient.

5

6-3

(continued)



Multiply 3 by 5. Write3 5 –6 4 –1 the result under –6.x 15

5 9 Add –6 and 15.

Step 3: Multiply the first coefficient by the new divisor. Write the result under the next coefficient. Add.

Step 4: Repeat the steps of multiplying and adding until the remainder is found.

The quotient is 5x2 + 9x + 31, R 92.

3 5 –6 4 –115 27 93

5 9 31 92

5x2 + 9x + 31 Remainder

6-3

The volume in cubic feet of a shipping carton is V(x) = x3 – 6x2 + 3x + 10. The height is x – 5 feet.



a. Find linear expressions for the other dimensions. Assume that the length is greater than the width.

b. If the width of the carton is 4 feet, what are the other two dimensions?

x2 – x – 2 = (x – 2)(x + 1) Factor the quotient.The length and the width are x + 1 and x – 2, respectively.

x – 2 = 4 Substitute 4 into the expression for width. Find x.x = 6Since the height equals x – 5 and the length equals x + 1, the height is 1 ft. and the length is 7 ft.

5 1 –6 3 10 Divide. 5 –5 –10

1 –1 –2 0

x2 – x – 2 Remainder

6-3

Use synthetic division to find P(3) for P(x) = x4 – 2x3 + x – 9.



By the Remainder Theorem, P(3) equals the remainder when P(x) is divided by x – 3.

The remainder is 21, so P(3) = 21.

3 1 –2 0 1 –9 3 3 9 30

1 1 3 10 21

6-3




1. x – 8

2. 3x – 5

3. x2 + 4x + 3, R 5

4. 2x2 + 5x + 2

5. 3x2 – 7x + 2

6. 9x – 12, R –32

7. x – 10, R 40

8. x2 + 4x + 3

9. no

10. yes

11. yes

12. no

13. x2 + 4x + 3

14. x2 – 2x + 2

15. x2 – 11x + 37, R –128

16. x2 + 2x + 5

17. x2 – x – 6

18. –2x2 + 9x – 19, R 40

19. x + 1, R 4

20. 3x2 + 8x – 3

21. x2 – 3x + 9

22. 6x – 2, R –4

23. y = (x + 1)(x + 3)(x – 2)

6-3



24. y = (x + 3)(x – 4)(x – 3)

25. = x + 3 and h = x

26. 18

27. 0

28. 0

29. 12

30. 168

31. 10

32. 51

33. 0

34. P(a) = 0; x – a is a factor of P(x).

35. x – 1 is not a factor of x3 – x2 – 2x because it does not divide into x3 – x2 – 2x evenly.

36. Answers may vary. Sample: (x2 + x – 4) ÷ (x – 2)

37. x2 + 4x + 5

38. x3 – 3x2 + 12x – 35, R 109

39. x4 – x3 + x2 – x + 1

40. x + 4

41. x3 – x2 + 1

42. no

43. yes

44. yes

6-3



45. no

46. no

47. yes

48. yes

49. yes

50. no

51. no

52. x3 – x2 + 1

53. x3 – 2x2 – 2x + 4, R –35

54. x3 – 2x2 – x + 6

55. x3 – 4x2 + x

56. a. x + 1

b. x2 + x + 1

c. x3 + x2 + x + 1

d. (x – 1)(x4 + x3 + x2 + x + 1)

57. a. x2 – x + 1

b. x4 – x3 + x2 – x + 1

c. x6 – x5 + x4 – x3 + x2 – x + 1

d. (x + 1)(x8 – x7 + x6 – x5 + x4 – x3 + x2 – x + 1)

58. By dividing it by a polynomial of degree 1, you are reducing the degree-n polynomial by one, to n – 1. The remainder will be constant because it is not divisible by the variable.

59. x + 2i

6-3



60. Yes; the graph could rise to the right and fall to the left or it could fall tothe right and rise to the left.

61. D

62. I

63. B

64. [2] x2 – 6x – 7x – 1 x3 – 7x2 – x + 7 x3 – x2

–6x2 – x –6x2 + 6x

– 7x + 7 – 7x + 7

0

x2 – 6x – 7 = 0 (x + 1)(x – 7) = 0 x = 1, –1, 7

[1] incorrect illustration of division or incorrect list of zeros

65. y = x2 + 2x – 15

66. y = x3 –9x2 + 8x

67. y = x3 – 6x2 + 3x + 10

68. y = x4 – 4x3 + 6x2 – 4x + 1

69. 24

70. 5

71. 23 – 11i

6-3



72. –0.5 0 –1.50.5 1 1.50.5 0 0.5

73. none exists

74. 0 2 11 –4 –20.5 –3.5 –1.5

6-3



1a. Divide using long division.(9x3 – 48x2 + 13x + 3) ÷ (x – 5)

1b. Is x – 5 a factor of 9x3 – 48x2 + 13x + 3?

2. Divide using synthetic division.(6x3 – 4x2 + 14x – 8) ÷ (x + 2)

3. Use synthetic division and the given factor x – 4 to completely factor x3 – 37x + 84.

4. Use synthetic division and the Remainder Theorem to findP(–2) when P(x) = x4 – 2x3 + 4x2 + x + 1.

9x2 – 3x – 2, R –7

no

6x2 – 16x + 46, R –100

(x + 7)(x – 3)(x – 4)

47

6-3



Solving Polynomial EquationsSolving Polynomial Equations

Graph each system. Find any points of intersection.

1. 2. 3.

Factor each expression.

4. x2 – 2x – 15 5. x2 – 9x + 14 6. x2 + 6x + 5

y = 3x + 1y = –2x + 6

–2x + 3y = 0 x + 3y = 3

2y = – x + 8x + 2y = –6

6-4

Solutions



1. 2.

intersection: (1, 4) 3. Rewrite equations:

Rewrite equations:

intersection: (1, ) no points of intersection4. Factors of –15 with a sum of –2: –5 and 3

x2 – 2x – 15 = (x – 5)(x + 3)5. Factors of 14 with a sum of –9: –7 and –2

x2 – 9x + 14 = (x – 7)(x – 2)6. Factors of 5 with a sum of 6: 5 and 1

x2 + 6x + 5 = (x + 5)(x + 1)

y = 3x + 1y = –2x + 6

–2x + 3y = 0 x + 3y = 3

2y = – x + 8x + 2y = –6 y = x

y = – x + 1

23

13

y = – x + 4

y = – x – 3

1212

6-4

23

Graph and solve x3 – 19x = –2x2 + 20.



Step 2: Use the Intersect feature to find thex values at the points of intersection.

Step 1: Graph y1 = x3 – 19x and y2 = –2x2 + 20 on a graphing calculator.

The solutions are –5, –1, and 4.

6-4

(continued)



Check: Show that each solution makes the original equation a true statement.

x3 – 19x = –2x2 + 20 x3 – 19x = –2x2 + 20 (–5)3 – 19(–5) –2(–5)2 + 20 (–1)3 – 19(–1) –2(–1)2 + 20

–125 + 95 –50 + 20 –1 + 19 –2 + 20–30 = –30 18 = 18

x3 – 19x = –2x2 + 20(4)3 – 19(4) –2(4)2 + 20

64 – 76 –32 + 20–12 = –12

6-4

The dimensions in inches of the cubicle area inside a doghouse

can be expressed as width x, length x + 4, and height x – 3. The volume

is 15.9 ft3. Find the dimensions of the doghouse.



V = l • w • h Write the formula for volume.

27475.2 = (x + 4)x(x – 3) Substitute.

15.9 ft3 • = 27475.2 in.3 Convert the volume to cubic inches.123 in.3

ft3

The dimensions of the doghouse are about 30 in. by 27 in. by 34 in.

6-4

Graph y1 = 27475.2 and y2 = (x + 4)x(x – 3). Use the Intersect option of the calculator.When y = 27475.2, x 30. So x – 3 27and x + 4 34.

Factor x3 – 125.



x3 – 125 = (x)3 – (5)3 Rewrite the expression as the difference of cubes.

= (x – 5)(x2 + 5x + (5)2) Factor.

= (x – 5)(x2 + 5x + 25) Simplify.

6-4

Solve 8x3 + 125 = 0. Find all complex roots.



8x3 + 125 = (2x)3 + (5)3 Rewrite the expression as the difference of cubes.

= (2x + 5)((2x)2 – 10x + (5)2) Factor.

= (2x + 5)(4x2 – 10x + 25) Simplify.

The quadratic expression 4x2 – 10x + 25 cannot be factored, so use the Quadratic Formula to solve the related quadratic equation 4x2 – 10x + 25 = 0.

6-4

Since 2x + 5 is a factor, x = – is a root.52

(continued)



x = Quadratic Formula

= Substitute 4 for a, –10 for b, and 25 for c.

–b ± b2 – 4ac2a

–(–10) ± (–10)2 – 4(4)(25)2(4)

= Use the Order of Operations. – (–10) ± –300

8

6-4

= –1 = 110 ± 10i 3

8

= Simplify.5 ± 5i 3

4

The solutions are – and .52

5 ± 5i 34

Factor x4 – 6x2 – 27.



Step 1: Since x4 – 6x2 – 27 has the form of a quadratic expression, you can factor it like one. Make a temporary substitution of variables.

x4 – 6x2 – 27 = (x2)2 – 6(x2) – 27 Rewrite in the form of a quadratic expression.

= a2 – 6a – 27 Substitute a for x2.

Step 2: Factor a2 – 6a – 27.

a2 – 6a – 27 = (a + 3)(a – 9)

Step 3: Substitute back to the original variables.

(a + 3)(a – 9) = (x2 + 3)(x2 – 9) Substitute x2 for a. = (x2 + 3)(x + 3)(x – 3) Factor completely.

The factored form of x4 – 6x2 – 27 is (x2 + 3)(x + 3)(x – 3).

6-4

Solve x4 – 4x2 – 45 = 0.



x4 – 4x2 – 45 = 0

(x2)2 – 4(x2) – 45 = 0 Write in the form of a quadratic expression. Think of the expression as a2 – 4a – 45, which factors as (a – 9)(a + 5).

(x2 – 9)(x2 + 5) = 0

(x – 3)(x + 3)(x2 + 5) = 0

x = 3 or x = –3 or x2 = –5 Use the factor theorem.

x = ± 3 or x = ± i 5 Solve for x, and simplify.

6-4




1. –2, 1, 5

2. –1, 0, 3

3. 0, 1

4. 0, 8

5. 0, –1, –2

6. 0, –3.5, 1

7. 0, –0.5, 1.5

8. –0.5, 0, 3

9. 1, 7

10. 4.8%

11. about 5.78 ft 6.78 ft 1.78 ft

12. (x + 4)(x2 – 4x + 16)

13. (x – 10)(x2 + 10x + 100)

14. (5x – 3)(25x2 + 15x + 9)

15. 3,

16. –4, 2 ± 2i 3

17. 5,

18. –1,

19. ,

20. – ,

21. (x2 – 7)(x – 1)(x + 1)

22. (x2 + 10)(x2 – 2)

–3 ± 3i 32

–5 ± 5i 32

1 ± i 32

1 ± i 34

12

12

–1 ± i 34

6-4



23. (x2 – 3)(x – 2)(x + 2)

24. (x – 2)(x + 2)(x – 1)(x + 1)

25. (x – 1)(x + 1)(x2 + 1)

26. 2(2x2 – 1)(x + 1)(x – 1)

27. ±3, ±1

28. ±2

29. ±4, ±2i

30. ±3i, ± 2

31. ± 2, ±i 6

32. ±i 5, ±i 3

33. –1, 3.24, –1.24

34. –9, 0

35. –2, –3, 1, 2

36. 1.71, 0.83

37. 0, 1.54, 8.46

38. 0, 1.27, 4.73

39. –1.04, 0, 6.04

40. (n – 1)(n)(n + 1) = 210; 5, 6, 7

41. about 3.58 cm, about 2.83 cm

42. – ,

43. ,

44. ±2 2, ±2i 2

45. ±5, ±i 2

3 ± 3i 35

–2 ± 2i 33

65

43

6-4



46. ±3i, ±i 3

47. 0, ±2, ±1

48. ± 10, ±i 10

49. 0, ±

50. 4, –2 ± 2i 3

51. 0, 3 ± 3

52. – , 0, 4

53. –1, 1, ±i 5

54. –3, –2, 2

55. –1, 3, 3

56. 0, 1, 3

57. 0, 0, 1, 6

32

12

26510

58. ± , ±i

59. ± 2, ±i

60. Check students’ work.

61. V = x2(4x – 2), 4 in. by 4 in. by 16 in.

62. x = length, V = x(x – 1)(x – 2), 5 meters

63. – , 1; y = (2x + 5)(x – 1)

64. ±3, ±1; y = (x – 1)(x + 1)(x – 3)(x + 3)

65. –1, 2, 2; y = (x + 1)(x – 2)2

66. –2, 1, 3; y = (x + 2)(x – 1)(x – 3)

67. –4, –1, 3; y = (x + 4)(x + 1)(x – 3)

68. A cubic can only have 3 zeros.

32

52

6-4



69. a. Answers may vary. Sample: x4 – 9 = 0, ± 3, ±i 3

b. No; two of the roots are imaginary.

70. Answers may vary. Sample: The pink block has volume a2(a – 3), the orange block has volume 9(a – 3), the blue block has volume 3a(a - 3), and the purple block has volume 27. Thus a3 – 27 = a2(a – 3) + 3a(a – 3) + 9(a – 3) = (a2 + 3a + 9)(a – 3).

71. a. 10b. 8 and 12

72. A

73. H74. [2] + = +

[1] attempts to write each of the terms as a cube

75. [4] 8x3 – 27 = (2x)3 – (3)3 = (2x – 3)(4x2 + 6x + 9).

If 2x – 3 = 0, then x = .

If 4x2 + 6x + 9 = 0,

then x = =

=

= .[3] minor computational errors[2] (2x – 3) = 0 and 4x2 + 6x + 9 = 0

written and solved for x in the first equation but not in the second

[1] answer correct, without work shown

–6 ± 6i 38

32

–3 ± 3i 34

–6 ± 36 – 1448

–6 ± 1088

18

a3

b6

a b2

3 312

6-4



76. x2 – 3x – 10

77. 2x2 + x – 3, R 2

78. –2, 6

79. ±6

80. – , 3

81. 0

–5

82. no unique solution

12

6-4



1. Solve x3 – 2x2 – 3 = x – 4 by graphing. Where necessary, round to the nearest hundredth.

Factor each expression.

2. 216x3 – 1

3. 8x3 + 125

4. x4 – 5x2 + 4

Solve each equation.

5. x3 + 125 = 0 6. x4 + 3x2 – 28 = 0

–0.80, 0.55, 2.25

(6x – 1)(36x2 + 6x + 1)

(2x + 5)(4x2 – 10x + 25)

(x + 1)(x – 1)(x + 2)(x – 2)

±2, ±i 7

6-4

5 ± 5i 34

–5,


(For help, go to Lessons 1-1, 5-1, and 5-6.)

Theorems About Roots of Polynomial EquationsTheorems About Roots of Polynomial Equations

List all the factors of each number.

1. 12 2. 24 3. 36 4. 48

Multiply.

5. (x – 5)(x2 + 7) 6. (x + 2)(x + 3)(x – 3)

Define each set of numbers.

7. rational 8. irrational 9. imaginary

6-5



Solutions

1. Factors of 12: ±1, ±2, ±3, ±4, ±6, ±12

2. Factors of 24: ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24

3. Factors of 36: ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36

4. Factors of 48: ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, ±48

5. (x – 5)(x2 + 7) = x3 + 7x – 5x2 – 35

= x3 – 5x2 + 7x – 35

6. (x + 2)(x + 3)(x – 3) = (x + 2)(x – ( 3)2) = (x + 2)(x2 – 3) =

x3 – 3x + 2x2 – 6 = x3 + 2x2 – 3x – 6

7. Rational numbers are all the numbers that can be written as the quotient of

two integers: , where b = 0.

8. Irrational numbers are numbers that cannot be written as a quotient of integers.

9. An imaginary number is any number of the form a + bi, where b = 0 and i is

defined as the number whose square is –1.

ab /

/

6-5

Find the rational roots of 3x3 – x2 – 15x + 5.



Step 1: List the possible rational roots.

6-5

The leading coefficient is 3. The constant term is 5. By the Rational

Root Theorem, the only possible rational roots of the equation have

the form .factors of 5factors of 3

The factors of 5 are ±5 and ±1 and ±5. The factors of 3 are

±3 and ±1. The only possible rational roots are ±5, ± , ±1, ± .53

13

(continued)



Step 2: Test each possible rational root.

5: 3(5)3 – (5)2 – 15(5) + 5 = 280 = 0–5: 3(–5)3 – (–5)2 – 15(–5) + 5 = –320 = 0

//

1: 3(1)3 – (1)2 – 15(1) + 5 = –8 = 0–1: 3(–1)3 – (–1)2 – 15(–1) + 5 = 16 = 0

//

6-5

: 3 3 – 2 – 15 + 5 = –8.8 = 0

: 3 3 – 2 – 15 + 5 = –13.3 = 0

/

/

5353

( )53

( )53

( )53

– ( )53

– ( )53

– ( )53

–

13 ( )1

3– ( )1

3– ( )1

3–

: 3 3 – 2 – 15 + 5 = 0 So is a root.

: 3 3 – 2 – 15 + 5 = 9.7 = 0/

13

( )13

( )13

( )13

13

–

The only rational root of 3x3 – x2 – 15x + 5 = 0 is .13

Find the roots of 5x3 – 24x2 + 41x – 20 = 0.



Step 1: List the possible rational roots.

The leading coefficient is 5. The constant term is 20. By the

Rational Root Theorem, the only possible roots of the equation

have the form .factors of – 20factors of 5

The factors of –20 are ±1 and ±20, ±2 and ±10, and ±4 and ±5.

The only factors of 5 are ±1 and ±5. The only possible rational roots

are ± , ± , ± , ±1, ±2, ±4, ±5, ±10, and ±20.15

25

45

6-5

(continued)



Step 2: Test each possible rational root until you find a root.

Test : 5 3 – 24 2 ± 41 – 20 = –12.72 = 0

Test – : 5 3 – 24 2 ± 41 – 2 = –29.2 = 0

Test : 5 3 – 24 2 ± 41 – 20 = –7.12 = 0

Test – : 5 3 – 24 2 ± 41 – 20 = –40.56 = 0

Test : 5 3 – 24 2 ± 41 – 20 = 0 So is a root.

( )15

15

15

25

25

45

( )25

( )45

( )15

( )25

( )45

( )45

( )25

( )15

(– )15

(– )15

(– )15

(– )25

(– )25

(– )25

45

/

/

/

/

Step 3: Use synthetic division with the root you found in Step 2to find the quotient.

5 –24 41 –20 4 –16 205 –20 25 0

5x2 – 20x + 25 Remainder

45

6-5

(continued)



Step 4: Find the roots of 5x2 – 20x + 25 = 0.

5x2 – 20x + 25 = 0 5(x2 – 4x + 5) = 0 Factor out the GCF, 5. x2 – 4x + 5 = 0

x = Quadratic Formula = Substitute 1 for a, –4 for b,

and 5 for c.

–b ± b2 – 4ac2a

–(–4) ± (–4)2 – 4(1)(5)2(1)

= Use order of operations.

= –1 = i.

= 2 ± i Simplify.

4 ± –42

4 ± 2i 2

6-5

The roots of 5x3 – 24x2 + 41x – 20 = 0 are , 2 + i, and 2 – i.45

A polynomial equation with rational coefficients has the roots 2 – 5 and 7 . Find two additional roots.



By the Irrational Root Theorem, if 2 – 5 is a root, then its conjugate 2 + 5 is also a root.

If 7 is a root, then its conjugate – 7 also is a root.

6-5

A polynomial equation and real coefficients has the roots 2 + 9i with 7i. Find two additional roots.



By the Imaginary Root Theorem, if 2 + 9i is a root, then its complex conjugate 2 – 9i also is a root.

If 7i is a root, then its complex conjugate –7i also is a root.

6-5

Find a third degree polynomial with rational coefficients that has roots –2, and 2 – i.



Step 1: Find the other root using the Imaginary Root Theorem.

Since 2 – i is a root, then its complex conjugate 2 + i is a root.

Step 2: Write the factored form of the polynomial using the Factor Theorem.

(x + 2)(x – (2 – i))(x – (2 + i))

6-5

(continued)



Step 3: Multiply the factors.

(x + 2)[x2 – x(2 – i) – x(2 + i) + (2 – i)(2 + i)] Multiply (x – (2 – i))(x – (2 + i)).

(x + 2)(x2 – 2x + ix – 2x – ix + 4 – i 2) Simplify. (x + 2)(x2 – 2x – 2x + 4 + 1) (x + 2)(x2 – 4x + 5) Multiply. x3 – 2x2 – 3x + 10

A third-degree polynomial equation with rational coefficients and roots –2 and 2 – i is x3 – 2x2 – 3x + 10 = 0.

6-5



6-5

12

12

1 ± 32

1 ± 73

72

12. 1, –2,

13. – 5, 13

14. 4 + 6, – 3

15. 1 + 10, 2 – 2

16. 1 – i, 5i

17. 2 – 3i, –6i

18. 4 + i, 3 – 7i

19. x3 – x2 + 9x – 9 = 0

20. x3 + 3x2 – 8x + 10 = 0

21. x3 – 2x2 + 16x – 32 = 0

22. x3 – 3x2 – 8x + 30 = 0

23. x3 – 6x2 + 4x – 24 = 0


1. ±1, ±2; 1

2. ±1, ±2, ±3, ±6; 1, –2, –3

3. ±1, ±2, ±4; –1

4. ± , ±1, ±2, ±4, ±8; no rational roots

5. ±1, ±2, ±4, ±8, ±16; –2

6. ±1, ±3, ±5, ±15; no rational roots

7. 2, ±i 5

8. 5, ±i 7

9. –3, 1,

10. –5,

11. ± , ±3



6-5

24. x3 – x2 + 2 = 0

25. ± , ± , ± , ± , ± , ± , ± ,

±1, ± , ±2, ±3, ±6; , ,

26. ± , ± , ± , ± , ± , ±1, ±2, ± ,

±4, ±5, ±10, ±20; 2, ,

27. ± , ± , ± , ± , ± , ±1, ± , ±3,

± , ±7, ± , ±21; , – , 1, 3

28. ± , ± , ± , ± , ± , ± , ± , ± ,

± , ±1, ± , ± , ± , ±15,

±3, ±5; – , ,

1 12

16

14

12

29. x4 – 6x3 + 14x2 – 24x + 40 = 0

30. x4 – 2x3 – x2 + 6x – 6 = 0

31. x4 – 6x3 + 2x2 + 30x – 35 = 0

32. Never true; 5 is not a factor of 8, so by the Rational Root Theorem, 5 is not a root of the equation.

33. Sometimes true; since –2 is a factor of 8, –2 is a possible root of the equation.

34. Always true; use the Rational Root Theorem with p = a and q = 1.

35. Sometimes true; since 5 and – 5 are conjugates, they can be roots of a polynomial equation with integer coefficients.

13

23

34

32

12

32

23

1 10

15

25

12

45

52

25

52

73

16

12

13

76

32

72

212

13

72

14

54

12

34

18

58

38

154

52

32

158

152

12

32

52



36. Never true; since 2 + i and –2 – i are not conjugates, they cannot be the only imaginary roots of a polynomial equation with integer roots. If their conjugates were also roots, there would be four roots and the equation would have to be of fourth degree.

37. If 2i is a root, then so is –2i.

38. Answers may vary. Sample: x4 – x2 – 2 = 0; roots are ± 2 and ±i.

39. If b of a + b were rational, then the sum would also be rational. Thus the roots would be rational, which is not always the case.

40. a. 2 real, 2 imaginary; 4 imaginary; 4 real

b. 5 real; 3 real, 2 imaginary; 4 imaginary, 1 real

c. Answers may vary. Sample: It has an odd number of real solutions, but it must have at least one real solution.

41. Answers may vary. Sample: You cannot use the Irrational Root Theorem unless the equation has rational coefficients.

42. x2 + (–2 + i )x + 12 – 8i = 0

43. a–c. Answers may vary. Sample:a. x – 1 – 2 = 0b. x2 – 2(1 + 2 )x + (1 + 2 )2 = 0c. –1

6-5



6-5

44. D

45. H

46. [2] p is a factor of 3, so p = ±1 or ±3q is a factor of 2, so q = ±1 or ±2

[1] relates p and q to the coefficients or the constant of the polynomial

47. [4] The three factors are (x + 4), (x + 4i ), (x – 4i ). (x + 4)(x + 4i )(x – 4i ) = (x + 4)(x2 + 16) = x3 + 4x2 + 16x + 64. The leading coefficient is ,

so the equation is x3 + 6x2 + 24x + 96 = 0.

[3] leaves the answer as x3 + 4x2 + 16x + 64 = 0, OR makes minor errors

[2] identifies (x + 4), (x + 4i ), and (x – 4i ) as factors

[1] identifies the third root as 4i

48. – ,

49. ± 5, ±2i

50. ± 5, ±i 5

51. –1, 7

52. –2 ± 2i

53. ± 5

54. (1, –4)

55. (–6, 2)

56. (5, 0)

32

32

3 ± 3i 34

32

32



1. Use the Rational Root Theorem to list all possible rational roots of 3x3 + x2 – 15x – 5 = 0. Then find any actual rational roots.

2. Find the roots of 10x4 + x3 + 7x2 + x – 3 = 0.

3. A polynomial equation with rational coefficients has the roots 7 and –5i. Find two additional roots.

4. Find a third-degree polynomial equation with integer coefficients that has the roots 8 and 3i.

– 7, 5i

x3 – 8x2 + 9x – 72 = 0

±1, ±5, ± , ± ; –13

53

13

– , , ±i35

12

6-5



The Fundamental Theorem of AlgebraThe Fundamental Theorem of Algebra

Find the degree of each polynomial.

1. 3x2 – x + 5 2. –x + 3 – x3 3. –4x5 + 1

Solve each equation using the quadratic formula.

4. x2 + 16 = 0 5. x2 – 2x + 3 = 0 6. 2x2 + 5x + 4 = 0

6-6



Solutions

1. 3x2 – x + 5; degree 2

2. –x + 3 – x3; degree 3

3. –4x5 + 1; degree 5

4. x2 + 16 = 0

x = with a = 1, b = 0, and c = 16

x = = = = ± 4i

–b ± b2 – 4ac2a

– 0 ± 02 – 4(1)(16)2(1)

0 ± 0 – 642

± 8i 2

6-6



5. x2 – 2x + 3 = 0

x = with a = 1, b = –2, and c = 3

x = = = =

= = 1 ± i 2

6. 2x2 + 5x + 4 = 0

x = with a = 2, b = 5, and c = 4

x = = = =

–b ± b2 – 4ac2a

– (–2) ± (–2)2 – 4(1)(3)2(1)

2 ± 4 – 122

–b ± b2 – 4ac2a

2 ± – 82

2 ± (–1) • 4 • 22

2 ± 2i 22

– 5 ± 52 – 4(2)(4)2(2)

–5 ± 25 – 324

–5 ± – 74

–5 ± i – 74

Solutions (continued)

6-6

For the equation x4 – 3x3 + 4x + 1 = 0, find the number of

complex roots, the possible number of real roots, and the possible

rational roots.



By the corollary to the Fundamental Theorem of Algebra, x4 – 3x3 + 4x + 1 = 0 has four complex roots.

By the Imaginary Root Theorem, the equation has either no imaginary roots, two imaginary roots (one conjugate pair), or four imaginary roots (two conjugate pairs). So the equation has either zero real roots, two real roots, or four real roots.

By the Rational Root Theorem, the possible rational roots of the equation are ±1.

6-6

Find the number of complex zeros of ƒ(x) = x5 + 3x4 – x – 3. Find all the zeros.



By the corollary to the Fundamental Theorem of Algebra, there are five complex zeros. You can use synthetic division to find a rational zero.

Step 1: Find a rational root from the possible roots of ±1 and ±3. Use synthetic division to test each possible until you get a remainder of zero.

–3 1 3 0 0 –1 –3–3 0 0 0 3

1 0 0 0 –1 0

x4 –1

So –3 is one of the roots.

6-6

(continued)



Step 2: Factor the expression x4 – 1. x4 – 1 = (x2 – 1)(x2 + 1) Factor x2 – 1. = (x – 1)(x + 1)(x2 + 1) So 1 and –1 are also roots.

Step 3: Solve x2 + 1 = 0. x2 + 1 = 0

x2 = –1 x = ± iSo i and –i are also roots.

The polynomial function ƒ(x) = x5 + 3x4 – x – 3 has three real zeros of x = –3, x = 1, and x = –1, and two complex zeros of x = i, and x = –i.

6-6



6-6


1. 3 complex roots; number of real roots: 1 or 3 possible rational roots: ±1

2. 2 complex roots; number of real roots: 0 or 2possible rational roots: ± , ± , ±1, ±7

3. 4 complex roots; number of real roots: 0, 2, or 4possible rational roots: 0

4. 5 complex roots; number of real roots: 1, 3, or 5possible rational roots: ± , ±1, ± , ±5

5. 7 complex roots; number of real roots: 1, 3, 5, or 7possible rational roots: ±1, ±3

6. 1 complex root; number of real roots: 1possible rational roots: ± , ± , ±1, ±2, ±4, ±8

7. 6 complex roots; number of real roots: 0, 2, 4, or 6possible rational roots: ± , ±1, ± , ±7

8. 10 complex roots; number of real roots: 0, 2, 4, 6, 8, or 10possible rational roots: ±1

13

73

12

52

14

12

12

72



6-6

9. –1,

10. 3, ±i

11. 4,

12. 2, ± 3

13. ±2, ± 2

14. ±2, ±i

15. 0,

16. –6, ±i

17. 4 complex roots; number of real roots: 0, 2, or 4possible rational roots: ± , ±1, ±2, ± , ±13, ±26

1 ± i 74

1 ± i 32

3 ± 3 52

12

132

18. 5 complex roots; number of real roots: 1, 3, or 5possible rational roots: ±1, ±2, ±3, ±6, ±9, ±18

19. 3 complex roots; number of real roots: 1 or 3possible rational roots: ± , ± , ±1, ± , ±2, ±3, ±4, ±6, ±12

20. 6 complex roots; number of real roots: 0, 2, 4, or 6possible rational roots: ± , ± , ± , ±1, ± , ±2, ±3, ±4, ±6,

±8, ±12, ±24

21. 4, ±3i

22. –2, ± 5

13

23

43

14

12

34

32



6-6

23. –6,

24. – , –1 ± 2i

25. , ± 2i 5

26. ,

27. Answers may vary. Sample: y = x4 + 3x2 + 2

28. ±0.75

14

1225

–1 ± i 116

29. –3.24, 1.24

30. If you have no constant, then all terms have an x that can be factored out. The resulting expression will have a constant that can be used in the Rational Root Theorem.

31. Yes; for example, 2x2 – 11x + 5 has roots 0.5 and 5.

32. C

33. C

–1 ± i 2



6-6

34. A

35. C

36. x4 + 6x3 + 14x2 + 24x + 40 = 0

37. 3 (mult. 2)

38.

39.

–5 ± i 474

3 ± i 234



For each equation, state the number of complex roots, the possible number of real roots, and the possible rational roots.

1. x4 – 13x3 + x2 – 5 = 0 2. x7 – x5 + x2 – 3x + 3 = 0

Find all the zeros of each function.

3. x3 – 3x2 – 8x – 10 = 0 4. x3 + 3x2 – 2x – 6 = 0

5. x4 – 29x2 + 100 = 0

4; 0, 2, or 4; ±1, ±5 7; 1, 3, 5, or 7; ±1, ±3

5, –1±i –3, ± 2

±2, ±5

6-6



Permutations and CombinationsPermutations and Combinations

6-7

Simplify each expression.

1. 10 • 9 • 8 • 7 • 6 2. 3.

Let a b = 2a(a + b). Evaluate each expression.

4. 3 • 4 5. 2 • 7 6. 5 • 1 7. 6 • 10

4 • 3 • 26 • 5

7 • 6 • 5 • 4 • 3 • 2 • 14 • 3 • 2 • 1

*

Permutations and CombinationsPermutations and CombinationsALGEBRA 2 LESSON 6-7ALGEBRA 2 LESSON 6-7

Solutions

6-7

1. 10 • 9 • 8 • 7 • 6 = 30,240

2. =

3. = 7 • 6 • 5 = 210

4. 3 • 4 with a • b = 2a(a + b):

2(3)(3 + 4) = 2(3)(7) = 6(7) = 42

5. 2 • 7 with a • b = 2a(a + b):

2(2)(2 + 7) = 2(2)(9) = 4(9) = 36

6. 5 • 1 with a • b = 2a(a + b):

2(5)(5 + 1) = 2(5)(6) = 10(6) = 60

7. 6 • 10 with a • b = 2a(a + b):

2(6)(6 + 10) = 2(6)(16) = 12(16) = 192

4 • 3 • 26 • 5

7 • 6 • 5 • 4 • 3 • 2 • 14 • 3 • 2 • 1

/ // /

/ //

45

/ // /

In how many ways can 6 people line up from left to right for a

group photo?


Since everybody will be in the picture, you are using all the items from the original set. You can use the Multiplication Counting Principle or factorial notation.

There are six ways to select the first person in line, five ways to select the next person, and so on.

The total number of permutations is 6 • 5 • 4 • 3 • 2 • 1 = 6!.

6! = 720

The 6 people can line up in 720 different orders.

6-7

How many 4-letter codes can be made if no letter can be used twice?


Method 1: Use the Multiplication Counting Principle.26 • 25 • 24 • 23 = 358,800

Method 2: Use the permutation formula. Since there are 26 letters arranged 4 at a time, n = 26 and r = 4.

There are 358,800 possible arrangements of 4-letter codes with no duplicates.

26P4 = = = 358,800 26! (26 – 4)!

26!22!

6-7


Evaluate 10C4.

10C4 = 10! 4!(10 – 4)!

= 10! 4! • 6!

10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 14 • 3 • 2 • 1 • 6 • 5 • 4 • 3 • 2 • 1= / / / / / /

/ / / / / /

10 • 9 • 8 • 74 • 3 • 2 • 1=

= 210

6-7


A disk jockey wants to select 5 songs from a new CD that contains 12 songs. How many 5-song selections are possible?

Relate: 12 songs chosen 5 songs at a timeDefine: Let n = total number of songs.

Let r = number of songs chosen at a time.

Write: nCr = 12C5

You can choose five songs in 792 ways.

Use the nCr feature of your calculator.

6-7


A pizza menu allows you to select 4 toppings at no extra charge from a list of 9 possible toppings. In how many ways can you select 4 or fewer toppings?

The total number of ways to pick the toppings is 126 + 84 + 36 + 9 + 1 = 256.

There are 256 ways to order your pizza.

You may choose 4 toppings, 3 toppings, 2 toppings, 1 toppings, or none.

9C4 9C3 9C2 9C1 9C0

6-7


6-7


1. 120

2. 3,628,800

3. 6,227,020,800

4. 720

5. 665,280

6. 120

7. 120

8. 3003

9. a. 24b. 120

10. 8

11. 56

12. 336

13. 1680

14. 6

15. 120

16. 60,480

17. 60

18. 10,897,286,400

19. 4,151,347,200

20. 12

21. 15

22. 56

23. 1

24. 4

25. 35

26. 15

27. 35

28.

29. 4368

30. 21

31. 2600

32. 126

33. true because of the Comm. Prop. of Add.

5 18


6-7

34. true because of the Assoc. Prop. of Mult.

35. False; answers may vary. Sample: (3 + 2)! = 120 and 3! + 2! = 8

36. False; answers may vary. Sample: (3 2)! = 720 and 3! 2! = 12

37. False; answers may vary. Sample: (3!)! = 720 and (3!)2 = 36

38. False; answers may vary. Sample: (3!)2 = 36 and 3(2!) = 9

39. 3125

40. 60

41. 2 ways, because order matters

42. 0.000206

43. 84

44.

45. 5

46. permutation

47. permutation

48. combination

49. combination

50. 330,791,175

51. 210

52. 12,650

53. 5

4 19,393

1249


54. 9504

55. a. 56b. 56c. Answers may vary. Sample:

Each time you choose 3 of the 8 points to use as vertices of a ,

the 5 remaining points could be used to form a pentagon.

56. 120

57. 3024

58. 360

59. 24

60. 1680

61. 840

62. 5040

63. 0

64–67. Check students’ work.

68. a. 2048b. Answers may vary. Sample:

No, because there are too many possible solutions.

69. a. The graph for y = xCx–2 is identical to the graph for y = xC2 because 2C2–2 = 2C2,

3C3–2 = 3C2, 4C4–2 = 4C2,

5C5–2 = 5C2, etc.

b. Answers may vary. Sample: The function is defined only at discrete whole-number values of x, and not over a smooth range of points as in a continuous function.

6-7


6-7

70. a. 35b. 6c. 7C3 = , so 7C3 • 3! = ,

which is the permutation formula for 7P3.

71. a. All the terms contain the factors 2 and 5. Since multiplication is

commutative, 2 5 = 10 and 10 times any integer ends in zero.

b. 24 zeros

72. Answers may vary. Sample: 99

73. a. 18 peopleb. 8568c. 658,008d. 0.013 or 1.3%

7! 3!4!

7!4!

74. 5040

75. 1/336

76. 21

77. 118

78. 351

79. 47/60

80. 600

81. 3 complex roots; number of real roots: 1 or 3possible rational roots: ± , ± , ± , ± , ± , ± , ±1, ±2

82. 2, ±i 6

83. –x3 – 3x2 + 6; cubic trinomial

1 12

16

14

13

12

23


84. 2x2 – 4x + 8; quadratic trinomial

85. 5t 2 – 3t; quadratic binomial

86. x4 – 100; quartic binomial

87. x3 – 4x; cubic binomial

88. t 4 – 2t 3 + t 2; quartic trinomial

89. 4(x – 1)2

90. –(x + 3)2

91. 3(x – 5)(x + 5)

92. minimum; –12

93. maximum; 4.125

94. minimum; –7

6-7


Evaluate each expression.

1. 8! 2. 3. 7C3 4. 9P5

5. There are 12 students waiting to enter a van to go on a trip to the science museum. In how many orders can they enter the van?

6. You have 8 shopping trips to make this week. You have time to make any 4 of them today. In how many ways can you select the four that you make today?

7. You want to hang 6 pictures in a row on a wall. You have 11 pictures from which to choose. How many picture arrangements are possible?

14!10!4!

40,320 1001 35 15,120

479,001,600 orders

70 ways

332,640 arrangements

6-7



The Binomial TheoremThe Binomial Theorem

Multiply.

1. (x + 2)2 2. (2x + 3)2 3. (x – 3)3 4. (a + b)4

Evaluate.

5. 5C0 6. 5C1 7. 5C2 8. 5C3 9. 5C4

6-8

The Binomial TheoremThe Binomial TheoremALGEBRA 2 LESSON 6-8ALGEBRA 2 LESSON 6-8

Solutions

1. (x + 2)2 = x2 + 2(2)x + 22 = x2 + 4x + 4

2. (2x + 3)2 = (2x)2 + 2(3)(2x) + 32 = 4x2 + 12x + 9

3. (x – 3)3 = (x – 3)(x – 3)2 = (x – 3)(x2 – 2(3)x + 32) =(x – 3)(x2 – 6x + 9) = x(x2 – 6x + 9) – 3(x2 – 6x + 9) =x3 – 6x2 + 9x – 3x2 + 18x – 27 = x3 – 9x2 + 27x – 27

4. (a + b)4 = (a + b)2(a + b)2 = (a2 + 2ab + b2)(a2 + 2ab + b2) =a2(a2 + 2ab + b2) + 2ab(a2 + 2ab + b2) + b2(a2 + 2ab + b2) =a4 + 2a3b + a2b2 + 2a3b + 4a2b2 + 2ab3 + a2b2 + 2ab3 + b4 =a4 + 4a3b + 6a2b2 + 4ab3 + b4

6-8


6-8

Solutions (continued)

5. 5C0 = = = = 1

6. 5C1 = = = = = 5

7. 5C2 = = = = = 10

8. 5C3 = = = = = 10

9. 5C4 = = = = = 5

5 • 4 • 3 • 2 • 12 • 1 • 3 • 2 • 1

5! 0! (5 – 0)!

5! 2! (5 – 2)!

5! 1! (5 – 1)!

5! 3! (5 – 3)!

5! 4! (5 – 4)!

5! 1 • 5!

5! 1! • 4!

5! 2! • 3!

5! 3! • 2!

5! 4! • 1!

11

5 • 4 • 3 • 2 • 11 • 4 • 3 • 2 • 1

/ / / /

/ / /

51

/ / /202

202

51

5 • 4 • 3 • 2 • 13 • 2 • 1 • 2 • 1

5 • 4 • 3 • 2 • 14 • 3 • 2 • 1 • 1

/ / // / /

/ / / /

/ / / // / / /

//

Use Pascal’s Triangle to expand (a + b)5.


Use the row that has 5 as its second number.

In its simplest form, the expansion is a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5.

The exponents for a begin with 5 and decrease.

1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + 1a0b5

The exponents for b begin with 0 and increase.

6-8

Use Pascal’s Triangle to expand (x – 3)4.


First write the pattern for raising a binomial to the fourth power.

1 4 6 4 1 Coefficients from Pascal’s Triangle.

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

Since (x – 3)4 = (x + (–3))4, substitute x for a and –3 for b.

(x + (–3))4 = x4 + 4x3(–3) + 6x2(–3)2 + 4x(–3)3 + (–3)4

= x4 – 12x3 + 54x2 – 108x + 81

The expansion of (x – 3)4 is x4 – 12x3 + 54x2 – 108x + 81.

6-8


Use the Binomial Theorem to expand (x – y)9.


Write the pattern for raising a binomial to the ninth power.(a + b)9 = 9C0a9 + 9C1a8b + 9C2a7b2 + 9C3a6b3

+ 9C4a5b4 + 9C5a4b5 + 9C6a3b6 + 9C7a2b7 + 9C8ab8 + 9C9b9

Substitute x for a and –y for b. Evaluate each combination.(x – y)9 = 9C0x9 + 9C1x8(–y) + 9C2x7(–y)2 + 9C3x6(–y)3

+ 9C4x5(–y)4 + 9C5x4(–y)5 + 9C6x3(–y)6 + 9C7x2(–y)7 + 9C8x(–y)8 + 9C9(–y)9

= x9 – 9x8y + 36x7y2 – 84x6y3 + 126x5y4 – 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9

The expansion of (x – y)9 is x9 – 9x8y + 36x7y2 – 84x6y3 + 126x5y4

– 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9.

6-8


Dawn Staley makes about 90% of the free throws she attempts. Find the probability that Dawn makes exactly 7 out of 12 consecutive free throws.


Since you want 7 successes (and 5 failures), use the term p7q5. This term has the coefficient 12C5.

Probability (7 out of 10) = 12C5 p7q5

= 0.0037881114 Simplify.

Dawn Staley has about a 0.4% chance of making exactly 7 out of 12 consecutive free throws.

= • (0.9)7(0.1)5 The probability p of success = 90%, or 0.9. 12! 5! •7!

6-8



1. a3 + 3a2b + 3ab2 + b3

2. x2 – 2xy + y2

3. a4 + 4a3b + 6a2b2 + 4ab3 + b4

4. x5 – 5x4y + 10x3y2 – 10x2y3 + 5xy4 – y5

5. a6 – 6a5b + 15a4b2 – 20a3b3 + 15a2b4 – 6ab5 + b6

6. x7 – 7x6y + 21x5y2 – 35x4y3 + 35x3y4 – 21x2y5 + 7xy6 – y7

7. x8 + 8x7y + 28x6y2 + 56x5y3 + 70x4y4 + 56x3y5 + 28x2y6 + 8xy7 + y8

8. d9 + 9d8e + 36d7e2 + 84d6e3 + 126d5e4 + 126d4e5 + 84d3e6 + 36d2e7 + 9de8 + e9

9. x3 – 9x2 + 27x – 27

10. a4 + 12a3b + 54a2b2 + 108ab3 + 81b4

11. x6 – 12x5 + 60x4 – 160x3 + 240x2 – 192x + 64

12. x8 – 32x7 + 448x6 – 3584x5 + 17,920x4 – 57,344x3 + 114,688x2 – 131,072x + 65,536

13. x4 + 4x3y + 6x2y2 + 4xy3 + y4

14. w5 + 5w4 + 10w3 + 10w2 + 5w + 1

15. s2 – 2st + t 2

16. x6 – 6x5 + 15x4 – 20x3 + 15x2 – 6x + 1

6-8


17. x4 – 4x3y + 6x2y2 – 4xy3 + y4

18. p7 + 7p6q + 21p5q2 + 35p4q3 + 35p3q4 + 21p2q5 + 7pq6 + q7

19. x5 – 15x4 + 90x3 – 270x2 + 405x – 243

20. 64 – 48x + 12x2 – x3

21. a. about 25%b. about 21%c. about 12%

22. about 66%

23. x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7

24. x8 – 40x7y + 700x6y2 – 7000x5y3 + 43,750x4y4 – 175,000x3y5 + 437,500x2y6 – 625,000xy7 + 390,625y8

25. 81x4 – 108x3y + 54x2y2 – 12xy3 + y4

26. x5 – 20x4y + 160x3y2 – 640x2y3 + 1280xy4 – 1024y5

27. 117,649 – 201,684x + 144,060x2 – 54,880x3 + 11,760x4 – 1344x5 + 64x6

28. 8x3 + 36x2y + 54xy2 + 27y3

29. x4 + 2x2y2 + y4

30. x6 – 6x4y + 12x2y2 – 8y3

31. x6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1

32. x6 – 6x5 + 15x4 – 20x3 + 15x2 – 6x + 1

6-8


33. x5 + 10x4 + 40x3 + 80x2 + 80x + 32

34. x5 – 10x4 + 40x3 – 80x2 + 80x – 32

35. 16x4 + 96x3y + 216x2y2 +

216xy3 + 81y4

36. 27x3 + 135x2y + 225xy2 + 125y3

37. 64x6 + 384x5y + 960x4y2 + 1280x3y3 + 960x2y4 + 384xy5 + 64y6

38. 81x4 + 216x3y + 216x2y2 + 96xy3 + 16y4

39. 32x5 + 80x4y + 80x3y2 + 40x2y3 + 10xy4 + y5

40. 2187x7 + 5103x6y + 5103x5y2 + 2835x4y3 + 945x3y4 + 189x2y5 + 21xy6 + y7

41. x6 + 18x5y + 135x4y2 + 540x3y3 + 1215x2y4 + 1458xy5 + 729y6

42. x3 + 15x2y + 75xy2 + 125y3

43. a. about 31%b. about 16%c. about 16%

44. a. 6b. 84

45. 8C4x4y4

46. 9

47. 7, 7r 6s

48. 594x10

49. 80x2

6-8


50. 27x8

51. 264x10

52. x11

53. 64y6

54. –823,680x8y7

55. 314,928x7

56. 29,568x10y6

57. 1716x12y14

58. Answers may vary. Sample: Since one of the terms is negative and it is alternately raised to odd and even powers, the term is negative when raised to an odd power and positive when raised to an even power.

59. a. (s + 0.5)3

b. s3 + 1.5s2 + 0.75s + 0.125

60. The exponent of q should be 5 because the exponent of q should be the degree (7) minus the exponent of p.

61. 13, d12, 12d11e

62. 16, x15, –15x14y

63. 6, 32a5, 80a4b

64. 8, x7, –21x6y

65. a. –4b. (1 – i )4 = 1 – 4i – 6 + 4i + 1 = –4

66. (–1 + 3 • i )3 = –1 + 3i 3 + 9 – 3i 3 = 8

6-8


6-8

67. Answers may vary. Sample: A coin is tossed five times with the probability of heads on each toss 0.5. Write an expression for the probability of exactly 2 heads being tossed.

68. a. (k + 1)! = (k + 1) • (k) • (k - 1) • . . . • 1 = (k + 1)[(k) • (k - 1) • . . . • 1] = (k + 1) • k!

b. The derivation below finds a common denominator for the fractions that represent nCk and nCk+1, and then uses algebra to show that nCk + nCk+1 =

n+1Ck+1. In addition, the identity from part (a) is used three times.

nCk + nCk+1 = + = + =

= = = = n+1Ck+1

c. If you consider the row of Pascal’s Triangle containing just 1 to be row zero,

4C2 is 6, the third entry in the fourth row. 4C3 is 4, the fourth entry in the fourth row. 5C3 is 10, the fourth entry in the fifth row.

4C2 + 4C3 = 6 + 4 = 10 = 5C3

n! k!(n – k)!

n! (k + 1)!(n – k – 1)!

(k + 1)n! (k + 1)!(n – k)!

(n – k)n! (k + 1)!(n – k)!

(k + 1 + n – k)n!(k + 1)!(n – k)!

(n + 1)n! (k + 1)!(n – k)!

(n + 1)! (k + 1)!(n – k)!

(n + 1)! (k + 1)!((n + 1) – (k +1))!


69. D

70. I

71. D

72. G

73. [2] 7C1x6y1 or 7x6y[1] provides some expression in x

and y without the correct coefficients

74. [4] The sixth term is 7C5(2x)2(–3y)5 = 21(4x2)(–243y5) = –20,412x2y5

[3] incorrect coefficient but correct exponents for x and y

[2] 7C6(2x)2(–3y)5 correct, but not the rest of the answer

[1] presents some expression in x and y

75. 20

76. 24

77. 35

78. 39,916,800

79. 35

80. 20

81. 20.75; –12.60; –3, 1, 4

82. 65.67; –32.49; –5, –1, 5

83. y = (x – 3)2 – 7

84. y = (x + 3.5)2 – 13.25

85. y = –4(x – 0)2 + 9

6-8


Use Pascal’s Triangle to expand each binomial.

1. (x – y5)3

2. (2a + b)5

3. Use the Binomial Theorem to expand (3x – 2y)4.

4. What is the coefficient of a4b6 in the expansion of (a + b)10?

x3 – 3x2y5 + 3xy10 – y15

32a5 + 80a4b + 80a3b2 + 40a2b3 + 10ab4 + b5

81x4 – 216x3y + 216x2y2 – 96xy3 + 16y4

210

6-8

ALGEBRA 2 CHAPTER 6ALGEBRA 2 CHAPTER 6

Polynomials and Polynomial FunctionsPolynomials and Polynomial Functions

1. –8x4 + 3x2 + 9; quartic trinomial

2. 8x2 + x; quadratic binomial

3. 2x3 – 2x2 – 12x; cubic trinomial

4. t 3 – 3t – 2; cubic trinomial

5. –1.62, 0, 0.62

6. –5, –2, –1, 1

7. 0.65, –3.04

8. –1.26, 1

9. –0.73, 1, 2.73

6-A

38



21. , –4

22. x + 4

23. x2 + 5x – 15, R 24

24. 3x – 6, R 10

25. x2 – x + 3, R –20

26. 0

27. 0

28. 0

29. 84

30. 3

31. 720

32. 15

1 ± 52

6-A

10. y = x3 – x2 + x –

11. y = x4 + 2x3 – 3x2 – 6x

12. y = x3 + 12x2 + 48x + 64

13. y = x3 – x2 – x + 1

14. y = x4 – x2 – 2

15. y = x4 – 8x3 + 18x2 + 4x – 40

16. Answers may vary. Sample: Using ±i and ± 2, y = x4 – x2 – 2.

17. –2, – , ,

18. ± 3, –4, 1

19. – ,

20. 0, 1

185

195

65

23

78

32

23

–5 ± 212



33. 35

34. 20

35. 19,958,400

36. 9

37. 7

38. combination, 126

39. combination, 455

40. permutation, 120

41. x5 + 5x4z + 10x3z2 + 10x2z3 + 5xz4 + z5

42. 1 – 4t + 4t 2

43. 1.7%

44. Combinations can be used by starting with nC0 for the first term. Then the second term would be nC1, the third nC2, and so forth until you reach nCn–1, and finally nCn.

45. 2 + 2

6-A

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