techniques for factoring trinomialsexercise set a.1: factoring polynomials 708 university of houston...

APPENDICES

University of Houston Department of Mathematics 696

Appendices

Appendix A.1: Factoring Polynomials

Techniques for Factoring Trinomials

Techniques for Factoring Trinomials

Factorability Test for Trinomials:

Example:

Solution:

APPENDIX A.1 Factoring Polynomials

MATH 1330 Precalculus 697

Factoring Trinomials with Leading Coefficient 1:

APPENDICES




Example:

Solution:

APPENDICES


Factoring Trinomials with Leading Coefficient Different from 1:

APPENDICES


Example:

Solution:

Additional Example 1:

(a) 22 3 8x x

(b) 242 25 3x x



Solution:


Solution:

APPENDICES



Solution:




Solution:

APPENDICES



Solution:

Exercise Set A.1: Factoring Polynomials


At times, it can be difficult to tell whether or not a

quadratic of the form 2

ax bx c can be factored

into the form dx e fx g , where a, b, c, d, e, f,

and g are integers. If 2

4b ac is a perfect square, then

the quadratic can be factored in the above manner.

For each of the following problems,

(a) Compute 2

4b ac .

(b) Use the information from part (a) to

determine whether or not the quadratic can

be written as factors with integer coefficients.

(Do not factor; simply answer Yes or No.)

1. 2 5 3x x

2. 2 7 10x x

3. 2 6 16x x

4. 2 6 4x x

5. 29 x

6. 27x x

7. 22 7 4x x

8. 26 1x x

9. 22 2 5x x

10. 25 4 1x x

Factor the following polynomials. If the polynomial

can not be rewritten as factors with integer

coefficients, then write the original polynomial as your

answer.

11. 2 4 5x x

12. 2 9 14x x

13. 2 5 6x x

14. 2 6x x

15. 2 7 12x x

16. 2 8 15x x

17. 2 12 20x x

18. 2 7 18x x

19. 2 5 24x x

20. 2 9 36x x

21. 2 16 64x x

22. 2 6 9x x

23. 2 15 56x x

24. 2 6 27x x

25. 2 11 60x x

26. 2 19 48x x

27. 2 17 42x x

28. 2 12 64x x

29. 2 49x

30. 2 36x

31. 2 3x

32. 2 8x

33. 29 25x

34. 216 81x

35. 22 5 3x x

36. 23 16 15x x

37. 28 2 3x x

38. 24 16 15x x

39. 29 9 4x x

40. 25 17 6x x

41. 24 3 10x x

42. 29 21 10x x

43. 212 17 6x x

44. 28 26 7x x

Factor the following. Remember to first factor out the

Greatest Common Factor (GCF) of the terms of the

polynomial, and to factor out a negative if the leading

coefficient is negative.

45. 2 9x x

46. 2 16x x

47. 25 20x x

Exercise Set A.1: Factoring Polynomials


48. 24 28x x

49. 22 18x

50. 28 8x

51. 4 25 20x x

52. 33 75x x

53. 22 10 8x x

54. 23 12 63x x

55. 210 10 420x x

56. 24 40 100x x

57. 3 29 22x x x

58. 3 27 6x x x

59. 3 24 4x x x

60. 5 4 310 21x x x

61. 4 3 26 6x x x

62. 3 22 80x x x

63. 5 39 100x x

64. 12 1049 64x x

65. 250 55 15x x

66. 230 24 72x x

Factor the following polynomials. (Hint: Factor first

by grouping, and then continue to factor if possible.)

67. 3 22 25 50x x x

68. 3 23 4 12x x x

69. 3 25 4 20x x x

70. 3 29 18 25 50x x x

71. 3 24 36 9x x x

72. 3 29 27 4 12x x x

APPENDIX A.2 Dividing Polynomials


Appendix A.2: Dividing Polynomials

Polynomial Long Division and Synthetic Division

Polynomial Long Division and Synthetic Division

Long Division of Polynomials:

Example:

Solution:

APPENDICES


APPENDICES


Synthetic Division of Polynomials:

Example:

Solution:



A Comparison of Long Division and Synthetic Division

Let us now analyze the previous two examples, both of which solved the same problem using

long division and then synthetic division.

Long Division Synthetic Division

4 3 22 05 8 3 5x x x x x

Constant: 5

Change the sign of the constant term when

performing synthetic division.

4 3 22 05 8 3 5x x x x x

Notice the coefficients of the dividend: 2, 0, 8, 3, 5

Write the coefficients of the dividend (without

changing any signs). Do not forget the

‘placeholder’ for 30x .

Notice that the coefficients in each column of

the subtraction problems under the division

sign (at the left) are similar to the numbers in

each column of the synthetic division problem

(above). Remember that at the left, the signs

are changed when the expressions are

subtracted.

3 2

4 3 2

4 3

3 2

3 2

2

2

2 10 42 213

5 2 0 8 3 5

2

10 8

10

4

10

2 3

42

213 5

213

1

50

210

106

060

5

x x x

x x x x x

x x

x x

x x

x x

x x

x

x

10 50 210 1

5 | 2 0 8 3 5

2 10 42 213 1060

065

2 0 8| 55 3

| 2 0 3 55 8

APPENDICES


Notice that the numbers in the answer line of

the synthetic division problem are the same as

the coefficients of the quotient plus the final

remainder in the long division problem.

In the long division problem, there is one

column for each power of x, and the

arithmetic in each column is done with the

coefficients.

Synthetic division is a shortcut for doing the

arithmetic with the coefficients without having

to write down all the variables. Remember that

this synthetic division procedure ONLY works

when the divisor is of the form D x x c .

The Remainder Theorem:

3 2

4 3 2

4 3

3 2

3 2

2

2

5 2 0 8 3 5

2 10

10 8

10 50

42 3

42

2 10 42 213

1060

210

213 5

213 1065

x x x

x x x x x

x x

x x

x x

x x

x x

x

x

5 | 2 0 8 3 5

10 50 210 1

2 10 42 213 1 6

065

0 0




Solution:


Solution:

APPENDICES



Solution:




Solution:

APPENDICES





Solution:

Exercise Set A.2: Dividing Polynomials


Use long division to find the quotient and the

remainder.

1. 2 6 11

2

x x

x

2. 2 5 12

3

x x

x

3. 2 7 2

1

x x

x

4. 2 6 5

4

x x

x

5. 3 22 19 12

3

x x x

x

6. 5

33222 23

x

xxx

7. 12

12656 23

x

xxx

8. 3 212 13 22 14

3 4

x x x

x

9. 3 2

2

2 13 28 21

3 1

x x x

x x

10. 4 3 2

2

7 4 42 12

7 2

x x x x

x x

11. 64

144433222

2345

x

xxxx

12. 362

4282201024

23468

xx

xxxxx

13. 5

1532

34

x

xx

14. xx

xxx

2

72432

35

Use synthetic division to find the quotient and the

remainder.

15. 2 8 4

10

x x

x

16. 3

642

x

xx

17. 5

286133 23

x

xxx

18. 4

312 23

x

xxx

19. 1

43 24

x

xx

20. 1

8732 45

x

xxx

21. 5

101827113 234

x

xxxx

22. 2

1251832 234

x

xxxx

23. 2

83

x

x

24. 3

814

x

x

25. 21

3 574

x

xx

26. 31

234 9106

x

xxx

Evaluate P(c) using the following two methods:

(a) Substitute c into the function.

(b) Use synthetic division along with the

Remainder Theorem.

27. 2;254)( 23 cxxxxP

28. 1;3875)( 23 cxxxxP

Exercise Set A.2: Dividing Polynomials


29. 1;12487)( 23 cxxxxP

30. 3;14672)( 34 cxxxxP

Evaluate P(c) using synthetic division along with the

Remainder Theorem. (Notice that substitution without

a calculator would be quite tedious in these examples,

so synthetic division is particularly useful.)

31. 5;321703883)( 23567 cxxxxxxP

32. 2;11235103)( 2456 cxxxxxxP

33. 43234 ;12254)( cxxxxP

34. 273456 ;135932196)( cxxxxxxP

When the remainder is zero, the dividend can be

written as a product of two factors (the divisor and the

quotient), as shown below.

30

65 , so 30 5 6 .

2

62

3

x xx

x

, so 2

6 3 2x x x x

In the following examples, use either long division or

synthetic division to find the quotient, and then write

the dividend as a product of two factors.

35. 2 11 24

8

x x

x

36. 2 3 40

5

x x

x

37. 2 7 18

2

x x

x

38. 2 10 21

3

x x

x

39. 24 25 21

7

x x

x

40. 23 22 24

6

x x

x

41. 22 7 5

1

x x

x

42. 25 4 12

2

x x

x

APPENDICES


Appendix A.3: Geometric Formulas

Geometric Formulas

Geometric Formulas

The following two pages contain geometric formulas which may be helpful to you in this course.



s

h

r

h

w

r

h

b

h

b

1b

h

2b

r

d

s

w

Rectangle

Perimeter: 2 2P w

Area: A w

______________________________________________

Square

Perimeter: 4P s

Area: 2A s

______________________________________________

Parallelogram

Perimeter: Add the side lengths

Area: A bh

______________________________________________

Triangle


Area: 2

bhA

______________________________________________

Equilateral Triangle

Perimeter: 3P s

Area: 2 3

4

sA

______________________________________________

Circle

Circumference: 2C r d

Area: 2A r

______________________________________________

Trapezoid


Area: 1 2

2

b b hA

______________________________________________

Right Circular Cylinder

Lateral Area: 2L rh Ch

Total Surface Area: 2S L B ,

where B represents the area of

the base , so 22 2S rh r

Volume: 2V Bh r h

______________________________________________

Right Circular Cone

Lateral Area: 2

CL r

Total Surface Area: S L B ,


the base, so 2S r r

Volume: 2

3 3

Bh r hV

______________________________________________

Rectangular Prism

Lateral Area: 2 2L h wh Ph ,

where P represents the

perimeter of the base.

Total Surface Area: 2S L B ,


the base, so

2 2 2S h wh w

Volume: V Bh wh

______________________________________________

Sphere

Surface Area: 24S r

Volume: 34

3

rV

r

h



c a

b

45o

45o

x

x

2x

x

1 1,x y

2 2,x y y

,M x y

60o

30o

x

3x 2x

d

x 1 1,x y

2 2,x y y

Pythagorean Theorem

2 2 2a b c

______________________________________________

Distance Formula

Distance between the points 1, 1x y and 2 2,x y :

2 2

2 1 2 1d x x y y

______________________________________________

Midpoint Formula

Midpoint of the segment joining the points 1, 1x y and

2 2,x y :

1 2 1 2, ,2 2

x x y yM x y

______________________________________________

30o-60

o-90

o Triangle

In a 30o-60o-90o triangle, the length of

the hypotenuse is twice the length of

the shorter leg, and the length of the

longer leg is 3 times the length of

the shorter leg.

______________________________________________

45o-45

o-90

o Triangle

In a 45o-45o-90o triangle, the legs

are congruent, and the length of the

hypotenuse is 2 times the length

of either leg.

______________________________________________

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