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4 Factoring the Trinomial ax[Sup(2)] + bx + c with a ≠ 1

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Chapter 5 Factoring

E X A M P L E

1

The ac method

Factor each trinomial.

a) 2x2 ϩ 7x ϩ 6

b) 2x2 ϩ x Ϫ 6

c) 10x2 ϩ 13x Ϫ 3

Solution

a) In 2x2 ϩ 7x ϩ 6 we have a ϭ 2, b ϭ 7, and c ϭ 6. So,

ac ϭ 2 и 6 ϭ 12.

Now we need two integers with a product of 12 and a sum of 7. The pairs of

integers with a product of 12 are 1 and 12, 2 and 6, and 3 and 4. Only 3 and 4 have

a sum of 7. Replace 7x by 3x ϩ 4x and factor by grouping:

2x2 ϩ 7x ϩ 6 ϭ 2x2 ϩ 3x ϩ 4x ϩ 6

Replace 7x by 3x ϩ 4x.

ϭ (2x ϩ 3)x ϩ (2x ϩ 3)2 Factor out the common factors.

ϭ (2x ϩ 3)(x ϩ 2)

Factor out 2x ϩ 3.

Check by FOIL.

b) In 2x2 ϩ x Ϫ 6 we have a ϭ 2, b ϭ 1, and c ϭ Ϫ6. So,

ac ϭ 2(Ϫ6) ϭ Ϫ12.

Now we need two integers with a product of Ϫ12 and a sum of 1. We can list the

possible pairs of integers with a product of Ϫ12 as follows:

1 and Ϫ12

Ϫ1 and 12

2 and Ϫ6

3 and Ϫ4

Ϫ2 and 6

Ϫ3 and 4

Only Ϫ3 and 4 have a sum of 1. Replace x by Ϫ3x ϩ 4x and factor by grouping:

2x2 ϩ x Ϫ 6 ϭ 2x2 Ϫ 3x ϩ 4x Ϫ 6

Replace x by Ϫ3x ϩ 4x.

ϭ (2x Ϫ 3)x ϩ (2x Ϫ 3)2 Factor out the common factors.

ϭ (2x Ϫ 3)(x ϩ 2)

Factor out 2x Ϫ 3.

Check by FOIL.

c) Because ac ϭ 10(Ϫ3) ϭ Ϫ30, we need two integers with a product of Ϫ30 and a

sum of 13. The product is negative, so the integers must have opposite signs. We

can list all pairs of factors of Ϫ30 as follows:

1 and Ϫ30

Ϫ1 and 30

2 and Ϫ15

Ϫ2 and 15

3 and Ϫ10

Ϫ3 and 10

5 and Ϫ6

Ϫ5 and 6

The only pair that has a sum of 13 is Ϫ2 and 15:

10x2 ϩ 13x Ϫ 3 ϭ 10x2 Ϫ 2x ϩ 15x Ϫ 3

Replace 13x by Ϫ2x ϩ 15x.

ϭ (5x Ϫ 1)2x ϩ (5x Ϫ 1)3 Factor out the common factors.

ϭ (5x Ϫ 1)(2x ϩ 3)

Factor out 5x Ϫ 1.

Check by FOIL.

Now do Exercises 1–38

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E X A M P L E

5.4

2

Factoring the Trinomial ax2 ϩ bx ϩ c with a

1

349

Factoring a trinomial in two variables by the ac method

Factor 8x2 Ϫ 14xy ϩ 3y2

Solution

Since a ϭ 8, b ϭ Ϫ14, and c ϭ 3, we have ac ϭ 24. Two numbers with a product of

24 and a sum of Ϫ14 must both be negative. The possible pairs with a product of 24

follow:

Ϫ1 and Ϫ24

Ϫ3 and Ϫ8

Ϫ2 and Ϫ12

Ϫ4 and Ϫ6

Only Ϫ2 and Ϫ12 have a sum of Ϫ14. Replace Ϫ14xy by Ϫ2xy Ϫ 12xy and factor by

grouping:

8x2 Ϫ 14xy ϩ 3y2 ϭ 8x2 Ϫ 2xy Ϫ 12xy ϩ 3y2

ϭ (4x Ϫ y)2x ϩ (4x Ϫ y)(Ϫ3y)

ϭ (4x Ϫ y)(2x Ϫ 3y)

Check by FOIL.

Now do Exercises 39–44

U2V Trial and Error

After you have gained some experience at factoring by the ac method, you can often

find the factors without going through the steps of grouping. For example, consider

the polynomial

3x2 ϩ 7x Ϫ 6.

The factors of 3x2 can only be 3x and x. The factors of 6 could be 2 and 3 or 1 and 6.

We can list all of the possibilities that give the correct first and last terms, without

regard to the signs:

(3x

3)(x 2)

(3x 2)(x 3)

(3x 6)(x 1)

(3x 1)(x 6)

Because the factors of Ϫ6 have unlike signs, one binomial factor is a sum and the

other binomial is a difference. Now we try some products to see if we get a middle

term of 7x:

(3x ϩ 3)(x Ϫ 2) ϭ 3x2 Ϫ 3x Ϫ 6 Incorrect

(3x Ϫ 3)(x ϩ 2) ϭ 3x2 ϩ 3x Ϫ 6 Incorrect

If the trinomial has no common factor, then neither binomial factor can

have a common factor.

Actually, there is no need to try (3x 3)(x 2) or (3x 6)(x 1) because each contains

a binomial with a common factor. A common factor in the binomial causes a common

factor in the product. But 3x2 ϩ 7x Ϫ 6 has no common factor. So the factors must

come from either (3x 2)(x 3) or (3x 1)(x 6). So we try again:

(3x ϩ 2)(x Ϫ 3) ϭ 3x2 Ϫ 7x Ϫ 6 Incorrect

(3x Ϫ 2)(x ϩ 3) ϭ 3x2 ϩ 7x Ϫ 6 Correct

Even though there may be many possibilities in some factoring problems, it is

often possible to find the correct factors without writing down every possibility. We

can use a bit of guesswork in factoring trinomials. Try whichever possibility you think

might work. Check it by multiplying. If it is not right, then try again. That is why this

method is called trial and error.

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E X A M P L E

3

Trial and error

Factor each trinomial using trial and error.

a) 2x2 ϩ 5x Ϫ 3

b) 3x2 Ϫ 11x ϩ 6

Solution

a) Because 2x2 factors only as 2x и x and 3 factors only as 1 и 3, there are only two

possible ways to get the correct first and last terms, without regard to the signs:

The ac method is more systematic

than trial and error. However, trial

U

and error can beVfaster and easier,

especially

or second

trial

The

ac method

more

systematic

is correct.

than

trial and error. However, trial

and error can be faster and easier,

especially if your first or second trial

is correct.

(2x

1)(x 3)

and

(2x 3)(x 1)

Because the last term of the trinomial is negative, one of the missing signs must

be ϩ, and the other must be Ϫ. The trinomial is factored correctly as

2x2 ϩ 5x Ϫ 3 ϭ (2x Ϫ 1)(x ϩ 3).

Check by using FOIL.

b) There are four possible ways to factor 3x2 Ϫ 11x ϩ 6:

(3x

1)(x 6)

(3x 2)(x 3)

(3x

6)(x 1)

(3x 3)(x 2)

The first binomials of (3x 6)(x 1) and (3x 3)(x 2) have a common factor of 3.

Since there is no common factor in 3x2 Ϫ11x ϩ 6, we can rule out both of these possibilities. Since the last term in 3x2 Ϫ 11x ϩ 6 is positive and the middle term is negative, both signs in the factors must be negative. So the correct factorization is either

(3x Ϫ 1)(x Ϫ 6) or (3x Ϫ 2)(x Ϫ 3). By using FOIL we can verify that (3x Ϫ 2)(x Ϫ 3) ϭ

3x2 Ϫ 11x ϩ 6. So the polynomial is factored correctly as

3x2 Ϫ 11x ϩ 6 ϭ (3x Ϫ 2)(x Ϫ 3).

Now do Exercises 45–64

Factoring by trial and error is not just guessing. In fact, if the trinomial has a positive

leading coefficient, we can determine in advance whether its factors are sums or

differences.

Using Signs in Trial and Error

1. If the signs of the terms of a trinomial are ϩ ϩ ϩ, then both factors are

sums: x2 ϩ 5x ϩ 6 ϭ (x ϩ 2)(x ϩ 3).

2. If the signs are ϩ Ϫ ϩ, then both factors are differences: x2 Ϫ 5x ϩ 6 ϭ

(x Ϫ 2)(x Ϫ 3).

3. If the signs are ϩ ϩ Ϫ or ϩ Ϫ Ϫ, then one factor is a sum and the

other is a difference: x2 ϩ x Ϫ 6 ϭ (x ϩ 3)(x Ϫ 2) and x2 Ϫ x Ϫ 6 ϭ

(x Ϫ 3)(x ϩ 2).

In Example 4 we factor a trinomial that has two variables.

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E X A M P L E

5.4

4

Factoring the Trinomial ax2 ϩ bx ϩ c with a

1

351

Factoring a trinomial with two variables by trial and error

Factor 6x2 Ϫ 7xy ϩ 2y2.

Solution

We list the possible ways to factor the trinomial:

(3x

2y)(2x y)

(3x y)(2x 2y)

(6x 2y)(x y)

(6x y)(x 2y)

Note that there is a common factor 2 in (2x 2y) and in (6x 2y). Since there is no common factor of 2 in the original trinomial, the second and third possibilities will not work.

Because the last term of the trinomial is positive and the middle term is negative, both factors must contain subtraction symbols. Only the first possibility will give a middle term of

Ϫ7xy when subtraction symbols are used in both factors. So,

6x2 Ϫ 7xy ϩ 2y2 ϭ (3x Ϫ 2y)(2x Ϫ y).

Now do Exercises 65–74

U3V Factoring Completely

You can use the latest factoring technique along with the techniques that you learned

earlier to factor polynomials completely. Remember always to first factor out the

greatest common factor (if it is not 1).

E X A M P L E

5

Factoring completely

Factor each polynomial completely.

a) 4x3 ϩ 14x2 ϩ 6x

b) 12x2y ϩ 6xy ϩ 6y

Solution

a) 4x3 ϩ 14x2 ϩ 6x ϭ 2x(2x2 ϩ 7x ϩ 3)

ϭ 2x(2x ϩ 1)(x ϩ 3)

Factor out the GCF, 2x.

Factor 2x2 ϩ 7x ϩ 3.

Check by multiplying.

b) 12x 2y ϩ 6xy ϩ 6y ϭ 6y(2x2 ϩ x ϩ 1) Factor out the GCF, 6y.

To factor 2x 2 ϩ x ϩ 1 by the ac method, we need two numbers with a product of 2

and a sum of 1. Because there are no such numbers, 2x2 ϩ x ϩ 1 is prime and the

factorization is complete.

Now do Exercises 75–84

Our first step in factoring is to factor out the greatest common factor (if it is not 1).

If the first term of a polynomial has a negative coefficient, then it is better to factor out

the opposite of the GCF so that the resulting polynomial will have a positive leading

coefficient.

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E X A M P L E

6

Factoring out the opposite of the GCF

Factor each polynomial completely.

a) Ϫ18x3 ϩ 51x2 Ϫ 15x

b) Ϫ3a2 ϩ 2a ϩ 21

Solution

a) The GCF is 3x. Because the first term has a negative coefficient, we factor

out Ϫ3x:

Ϫ18x3 ϩ 51x2 Ϫ 15x ϭ Ϫ3x(6x2 Ϫ 17x ϩ 5)

Factor out Ϫ3x.

ϭ Ϫ3x(3x Ϫ 1)(2x Ϫ 5) Factor 6x2 Ϫ 17x ϩ 5.

b) The GCF for Ϫ3a2 ϩ 2a ϩ 21 is 1. Because the first term has a negative coefficient, factor out Ϫ1:

Ϫ3a2 ϩ 2a ϩ 21 ϭ Ϫ1(3a2 Ϫ 2a Ϫ 21) Factor out Ϫ1.

ϭ Ϫ1(3a ϩ 7)(a Ϫ 3)

Factor 3a2 Ϫ 2a Ϫ 21.

Now do Exercises 85–100

Warm-Ups

Fill in the blank.

5.4

1. If there are no two integers that have a

of ac

and a

of b, then ax2 ϩ bx ϩ c is prime.

2. In the

method we make educated

guesses at the factors and then check by FOIL.

True or false?

3.

4.

5.

6.

7.

8.

2x2 ϩ 3x ϩ 1 ϭ (2x ϩ 1)(x ϩ 1)

2x2 ϩ 5x ϩ 3 ϭ (2x ϩ 1)(x ϩ 3)

3x2 ϩ 10x ϩ 3 ϭ (3x ϩ 1)(x ϩ 3)

2x2 Ϫ 7x Ϫ 9 ϭ (2x Ϫ 9)(x ϩ 1)

2x2 Ϫ 16x Ϫ 9 ϭ (2x Ϫ 9)(2x ϩ 1)

12x2 Ϫ 13x ϩ 3 ϭ (3x Ϫ 1)(4x Ϫ 3)

Exercises

U Study Tips V

• Pay particular attention to the examples that your instructor works in class or presents to you online.

• The examples and homework assignments should give you a good idea of what your instructor expects from you.

U1V The ac Method

Find the following. See Example 1.

1. Two integers that have a product of 12 and a sum of 7

2. Two integers that have a product of 20 and a sum of 12

3. Two integers that have a product of 30 and a sum of Ϫ17

4. Two integers that have a product of 36 and a sum of Ϫ20

5. Two integers that have a product of Ϫ12 and a sum of Ϫ4

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5.4

6. Two integers that have a product of Ϫ8 and a sum

of 7

Each of the following trinomials is in the form ax2 ϩ bx ϩ c.

For each trinomial, find two integers that have a product of ac

and a sum of b. Do not factor the trinomials. See Example 1.

Factoring the Trinomial ax2 ϩ bx ϩ c with a

1

47. 6x2 ϩ 5x ϩ 1

48. 15y2 ϩ 8y ϩ 1

49. 5a2 ϩ 11a ϩ 2

50. 3y2 ϩ 10y ϩ 7

51. 4w2 ϩ 8w ϩ 3

52. 6z2 ϩ 13z ϩ 5

7. 6x2 ϩ 7x ϩ 2

8. 5x2 ϩ 17x ϩ 6

53. 15x2 Ϫ x Ϫ 2

54. 15x2 ϩ 13x Ϫ 2

9. 6y2 Ϫ 11y ϩ 3

10. 6z2 Ϫ 19z ϩ 10

55. 8x2 Ϫ 6x ϩ 1

56. 8x2 Ϫ 22x ϩ 5

11. 12w2 ϩ w Ϫ 1

12. 15t2 Ϫ 17t Ϫ 4

57. 15x2 Ϫ 31x ϩ 2

58. 15x2 ϩ 31x ϩ 2

353

Factor each trinomial using the ac method. See Example 1.

See the Strategy for Factoring ax2 ϩ bx ϩ c by the ac Method

box on page 347.

13. 2x2 ϩ 3x ϩ 1

14. 2x2 ϩ 11x ϩ 5

59. 4x2 Ϫ 4x ϩ 3

60. 4x2 ϩ 12x Ϫ 5

61. 2x2 ϩ 18x Ϫ 90

62. 3x2 ϩ 11x ϩ 10

63. 3x2 ϩ x Ϫ 10

64. 3x2 Ϫ 17x ϩ 10

15. 2x2 ϩ 9x ϩ 4

16. 2h2 ϩ 7h ϩ 3

65. 10x2 Ϫ 3xy Ϫ y2

66. 8x2 Ϫ 2xy Ϫ y2

17. 3t 2 ϩ 7t ϩ 2

18. 3t2 ϩ 8t ϩ 5

67. 42a2 Ϫ 13ab ϩ b2

68. 10a2 Ϫ 27ab ϩ 5b2

19. 2x2 ϩ 5x Ϫ 3

20. 3x2 Ϫ x Ϫ 2

21. 6x2 ϩ 7x Ϫ 3

22. 21x2 ϩ 2x Ϫ 3

23. 3x2 Ϫ 5x ϩ 4

25. 2x2 Ϫ 7x ϩ 6

24. 6x2 Ϫ 5x ϩ 3

26. 3a2 Ϫ 14a ϩ 15

27. 5b2 Ϫ 13b ϩ 6

28. 7y2 ϩ 16y Ϫ 15

29. 4y2 Ϫ 11y Ϫ 3

30. 35x2 Ϫ 2x Ϫ 1

31. 3x ϩ 2x ϩ 1

33. 8x2 Ϫ 2x Ϫ 1

32. 6x Ϫ 4x Ϫ 5

34. 8x2 Ϫ 10x Ϫ 3

Factor each polynomial completely. See Examples 5 and 6.

75. 81w3 Ϫ w

76. 81w3 Ϫ w2

35. 9t2 Ϫ 9t ϩ 2

36. 9t2 ϩ 5t Ϫ 4

77. 4w2 ϩ 2w Ϫ 30

78. 2x2 Ϫ 28x ϩ 98

37. 15x2 ϩ 13x ϩ 2

38. 15x2 Ϫ 7x Ϫ 2

79. 27 ϩ 12x2 ϩ 36x

80. 24y ϩ 12y2 ϩ 12

81. 6w2 Ϫ 11w Ϫ 35

82. 8y2 Ϫ 14y Ϫ 15

2

2

Use the ac method to factor each trinomial. See Example 2.

Complete the factoring.

69.

70.

71.

72.

73.

74.

3x2 ϩ 7x ϩ 2 ϭ (x ϩ 2)(

2x2 Ϫ x Ϫ 15 ϭ (x Ϫ 3)(

5x2 ϩ 11x ϩ 2 ϭ (5x ϩ 1)(

4x2 Ϫ 19x Ϫ 5 ϭ (4x ϩ 1)(

6a2 Ϫ 17a ϩ 5 ϭ (3a Ϫ 1)(

4b2 Ϫ 16b ϩ 15 ϭ (2b Ϫ 5)(

)

)

)

)

)

)

U3V Factoring Completely

39. 4a2 ϩ 16ab ϩ 15b2

40. 10x2 ϩ 17xy ϩ 3y2

83. 3x2z Ϫ 3zx Ϫ 18z

84. a2b ϩ 2ab Ϫ 15b

41. 6m2 Ϫ 7mn Ϫ 5n2

42. 3a2 ϩ 2ab Ϫ 21b2

85. 9x3 Ϫ 21x2 ϩ 18x

86. Ϫ8x3 ϩ 4x2 Ϫ 2x

43. 3x2 Ϫ 8xy ϩ 5y2

44. 3m2 Ϫ 13mn ϩ 12n2

87. a2 ϩ 2ab Ϫ 15b2

88. a2b2 Ϫ 2a2b Ϫ 15a2

89. 2x2y2 ϩ xy2 ϩ 3y2

90. 18x2 Ϫ 6x ϩ 6

91. Ϫ6t3 Ϫ t2 ϩ 2t

92. Ϫ36t2 Ϫ 6t ϩ 12

93. 12t4 Ϫ 2t3 Ϫ 4t 2

94. 12t3 ϩ 14t2 ϩ 4t

U2V Trial and Error

Factor each trinomial using trial and error. See Examples 3 and 4.

45. 5a2 ϩ 6a ϩ 1

46. 7b2 ϩ 8b ϩ 1

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95.

96.

97.

98.

99.

100.

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Chapter 5 Factoring

4x2y Ϫ 8xy2 ϩ 3y3

9x2 ϩ 24xy Ϫ 9y2

Ϫ4w2 ϩ 7w Ϫ 3

Ϫ30w2 ϩ w ϩ 1

Ϫ12a3 ϩ 22a 2b Ϫ 6ab2

Ϫ36a2b ϩ 21ab2 Ϫ 3b3

a) Rewrite the formula by factoring the right-hand side

completely.

b) Use the factored version of the formula to find N(3).

c) Use the accompanying graph to estimate the time at

which the workers are most efficient.

d) Use the accompanying graph to estimate the

maximum number of components assembled per

hour during an 8-hour shift.

Applications

Solve each problem.

101. Height of a ball. If a ball is thrown straight upward at

40 feet per second from a rooftop 24 feet above the

ground, then its height in feet above the ground t seconds

after it is thrown is given by

h(t) ϭ Ϫ16t2 ϩ 40t ϩ 24.

a) Find h(0), h(1), h(2), and h(3).

b) Rewrite the formula with the polynomial factored

completely.

c) Find h(3) using the result of part (b).

Getting More Involved

103. Exploration

Find all positive and negative integers b for which

each polynomial can be factored.

a) x2 ϩ bx ϩ 3

c) 2x2 ϩ bx Ϫ 15

b) 3x2 ϩ bx ϩ 5

104. Exploration

40 ft/sec

Find two integers c (positive or negative) for which

each polynomial can be factored. Many answers are

possible.

a) x 2 ϩ x ϩ c

b) x2 Ϫ 2x ϩ c

c) 2x 2 Ϫ 3x ϩ c

h(t) ϭ Ϫ16 t 2 ϩ 40t ϩ 24

105. Cooperative learning

Working in groups, cut two large squares, three

rectangles, and one small square out of paper that are

exactly the same size as shown in the accompanying

figure. Then try to place the six figures next to one

another so that they form a large rectangle. Do not

overlap the pieces or leave any gaps. Explain how

factoring 2x2 ϩ 3x ϩ 1 can help you solve this puzzle.

Figure for Exercise 101

102. Worker efficiency. In a study of worker efficiency at Wong

Laboratories it was found that the number of components

assembled per hour by the average worker t hours after

starting work could be modeled by the formula

N(t) ϭ Ϫ3t3 ϩ 23t2 ϩ 8t.

x

x

Number of components

300

1

1

1

1

1

x

x

x

x

x

200

Figure for Exercise 105

100

0

106. Cooperative learning

0 1 2 3 4 5 6 7 8

Time (hours)

Figure for Exercise 102

Working in groups, cut four squares and eight

rectangles out of paper as in Exercise 105 to illustrate the

trinomial 4x2 ϩ 7x ϩ 3. Select one group to demonstrate

how to arrange the 12 pieces to form a large rectangle.

Have another group explain how factoring the trinomial

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5.5

5.5

In This Section

U1V Factoring a Difference or Sum

of Two Cubes

2

U V Factoring a Difference of Two

Fourth Powers

3

U V The Factoring Strategy

Difference and Sum of Cubes and a Strategy

355

Difference and Sum of Cubes and a Strategy

In Sections 5.1 to 5.4, we established the general idea of factoring and some special

cases. In this section we will see two more special cases. We will then summarize all

of the factoring that we have done with a factoring strategy.

U1V Factoring a Difference or Sum of Two Cubes

We can use division to discover that a Ϫ b is a factor of a3 Ϫ b3 (a difference of two

cubes) and a ϩ b is a factor of a3 ϩ b3 (a sum of two cubes):

a2 ϩ ab ϩ b2

3 ෆෆෆෆ

ෆෆ

a Ϫ bͤෆaෆ

ϩ 0a2bෆ

ϩෆ0aෆbෆ2ෆϪ

bෆ3

3

2

a Ϫ ab

a2b ϩ 0ab2

a2b Ϫ ab2

ab2 Ϫ b3

ab2 Ϫ b3

0

a2 Ϫ ab ϩ b2

2 ෆෆෆෆ2ෆෆෆෆ

ͤ

a ϩ b ෆa3ෆෆ

ϩෆ0ෆaෆ

b ϩ 0ab ϩ b3

3

2

a ϩ ab

Ϫa2b ϩ 0ab2

Ϫa2b Ϫ ab2

ab2 ϩ b3

ab2 ϩ b3

0

In each division the remainder is 0. So in each case the dividend is equal to the divisor times the quotient. These results give us two new factoring rules.

Factoring a Difference or Sum of Two Cubes

a3 Ϫ b3 ϭ (a Ϫ b)(a2 ϩ ab ϩ b2)

a3 ϩ b3 ϭ (a ϩ b)(a2 Ϫ ab ϩ b2)

Use the following strategy to factor a difference or sum of two cubes.

Strategy for Factoring a3 ؊ b3 or a3 ؉ b3

1. The first factor is the original polynomial without the exponents, and the

middle term in the second factor has the opposite sign from the first factor:

a3 – b3 ϭ (a – b)(a2 ϩ ab ϩ b2)

opposite signs

a3 ϩ b3 ϭ (a ϩ b)(a2 – ab ϩ b2)

opposite signs

2. Recall the two perfect square trinomials a ϩ 2ab ϩ b2 and a2 – 2ab ϩ b2.

2

The second factor is almost a perfect square trinomial. Just delete the 2.

It is helpful also to compare the differences and sums of squares and cubes:

a2 Ϫ b2 ϭ (a Ϫ b)(a ϩ b)

a3 Ϫ b3 ϭ (a Ϫ b)(a2 ϩ ab ϩ b2)

a2 ϩ b2 Prime

a3 ϩ b3 ϭ (a ϩ b)(a2 Ϫ ab ϩ b2)

The factors a2 ϩ ab ϩ b2 and a2 Ϫ ab ϩ b2 are prime. They can’t be factored. The

perfect square trinomials a2 ϩ 2ab ϩ b2 and a2 Ϫ 2ab ϩ b2, which are almost the

same, are not prime. They can be factored:

a2 ϩ 2ab ϩ b2 ϭ (a ϩ b)2

and

a2 Ϫ 2ab ϩ b2 ϭ (a Ϫ b)2.

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Chapter 5 Factoring

E X A M P L E

1

Factoring a difference or sum of two cubes

Factor each polynomial.

a) w3 Ϫ 8

b) x3 ϩ 1

c) 8y3 Ϫ 27

Solution

a) Because 8 ϭ 23, w3 Ϫ 8 is a difference of two cubes. To factor w3 Ϫ 8, let

a ϭ w and b ϭ 2 in the formula a3 Ϫ b3 ϭ (a Ϫ b)(a2 ϩ ab ϩ b2):

w3 Ϫ 8 ϭ (w Ϫ 2)(w2 ϩ 2w ϩ 4)

b) Because 1 ϭ 13, the binomial x3 ϩ 1 is a sum of two cubes. Let a ϭ x and b ϭ 1 in

the formula a3 ϩ b3 ϭ (a ϩ b)(a2 Ϫ ab ϩ b2):

x3 ϩ 1 ϭ (x ϩ 1)(x2 Ϫ x ϩ 1)

c) 8y3 Ϫ 27 ϭ (2y)3 Ϫ 33

This is a difference of two cubes.

ϭ (2y Ϫ 3)(4y ϩ 6y ϩ 9) Let a ϭ 2y and b ϭ 3 in the formula.

2

Now do Exercises 1–16

In Example 1, we used the first three perfect cubes, 1, 8, and 27. You should verify

that 1, 8, 27, 64, 125, 216, 343, 512, 729, and 1000 are the first 10 perfect cubes.

CAUTION The polynomial (a Ϫ b)3 is not equivalent to a3 Ϫ b3 because if a ϭ 2

and b ϭ 1, then

(a Ϫ b)3 ϭ (2 Ϫ 1)3 ϭ 13 ϭ 1

and

a3 Ϫ b3 ϭ 23 Ϫ 13 ϭ 8 Ϫ 1 ϭ 7.

Likewise, (a ϩ b)3 is not equivalent to a3 ϩ b3.

U2V Factoring a Difference of Two Fourth Powers

A difference of two fourth powers of the form a4 Ϫ b4 is also a difference of two

squares, (a2)2 Ϫ (b2)2. It can be factored by the rule for factoring a difference of two

squares:

Write as a difference of two squares.

a4 Ϫ b4 ϭ (a2)2 Ϫ (b2)2

2

2

2

2

ϭ (a Ϫ b )(a ϩ b )

Difference of two squares

ϭ (a Ϫ b)(a ϩ b)(a2 ϩ b2) Factor completely.

Note that the sum of two squares a2 ϩ b2 is prime and cannot be factored.

E X A M P L E

2

Factoring a difference of two fourth powers

Factor each polynomial completely.

a) x4 Ϫ 16

b) 81m4 Ϫ n4

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Difference and Sum of Cubes and a Strategy

357

Solution

a) x4 Ϫ 16 ϭ (x2)2 Ϫ 42

Write as a difference of two squares.

ϭ (x Ϫ 4)(x ϩ 4)

2

2

Difference of two squares

ϭ (x Ϫ 2)(x ϩ 2)(x ϩ 4)

2

Factor completely.

b) 81m4 Ϫ n4 ϭ (9m2)2 Ϫ (n2)2

ϭ (9m Ϫ n

2

Write as a difference of two squares.

)(9m

2

2

ϩn

2

)

Factor.

ϭ (3m Ϫ n)(3m ϩ n)(9m2 ϩ n2)

Factor completely.

Now do Exercises 17–24

CAUTION A difference of two squares or cubes can be factored, and a sum of two

cubes can be factored. But the sums of two squares x2 ϩ 4 and 9m2 ϩ n2

in Example 2 are prime.

U3V The Factoring Strategy

The following is a summary of the ideas that we use to factor a polynomial completely.

Strategy for Factoring Polynomials Completely

1. Factor out the GCF (with a negative coefficient if necessary).

2. When factoring a binomial, check to see whether it is a difference of two

3.

4.

5.

6.

squares, a difference of two cubes, or a sum of two cubes. A sum of two

squares does not factor.

When factoring a trinomial, check to see whether it is a perfect square trinomial.

If the polynomial has four terms, try factoring by grouping.

When factoring a trinomial that is not a perfect square, use the ac method or

the trial-and-error method.

Check to see whether any of the factors can be factored again.

We will use the factoring strategy in Example 3.

E X A M P L E

3

Factoring polynomials

Factor each polynomial completely.

a) 2a2b Ϫ 24ab ϩ 72b

b) 3x3 ϩ 6x2 Ϫ 75x Ϫ 150

c) Ϫ3x4 Ϫ 15x3 ϩ 72x2

d) 60y3 Ϫ 85y2 Ϫ 25y

Solution

a) 2a2b Ϫ 24ab ϩ 72b ϭ 2b(a2 Ϫ 12a ϩ 36)

ϭ 2b(a Ϫ 6)2

First factor out the GCF, 2b.

Factor the perfect square

trinomial.

b) 3x3 ϩ 6x2 Ϫ 75x Ϫ 150 ϭ 3[x3 ϩ 2x 2 Ϫ 25x Ϫ 50] Factor out the GCF, 3.

ϭ 3[x 2(x ϩ 2) Ϫ 25(x ϩ 2)] Factor out common factors.

ϭ 3(x 2 Ϫ 25)(x ϩ 2)

Factor by grouping.

ϭ 3(x ϩ 5)(x Ϫ 5)(x ϩ 2)

Factor the difference

of two squares.

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Chapter 5 Factoring

c) Factor out Ϫ3x2 to get Ϫ3x4 Ϫ 15x3 ϩ 72x2 ϭ Ϫ3x2(x2 ϩ 5x Ϫ 24). To factor the

trinomial, find two numbers with a product of Ϫ24 and a sum of 5. For a product

of 24 we have 1 и 24, 2 и 12, 3 и 8, and 4 и 6. To get a sum of 5 and a product of

Ϫ24 choose 8 and Ϫ3:

Ϫ3x4 Ϫ 15x3 ϩ 72x2 ϭ Ϫ3x2(x2 ϩ 5x Ϫ 24)

ϭ Ϫ3x2(x Ϫ 3)(x ϩ 8)

d) Factor out 5y to get 60y3 Ϫ 85y2 Ϫ 25y ϭ 5y(12y2 Ϫ 17y Ϫ 5). By the ac method

we need two numbers that have a product of Ϫ60 (ac) and a sum of Ϫ17. The

numbers are Ϫ20 and 3. Now factor by grouping:

60y3 Ϫ 85y2 Ϫ 25y ϭ 5y(12y2 Ϫ 17y Ϫ 5)

ϭ 5y(12y2 Ϫ 20y ϩ 3y Ϫ 5)

ϭ 5y[4y(3y Ϫ 5) ϩ 1(3y Ϫ 5)]

ϭ 5y(3y Ϫ 5)(4y ϩ 1)

Factor out 5y.

Ϫ17y ϭ Ϫ20y ϩ 3y

Factor by grouping.

Factor out 3y Ϫ 5.

Now do Exercises 25–92

Warm-Ups

Fill in the blank.

5.5

1. If there is no

, then the dividend is the divisor

times the quotient.

2. The binomial a3 ϩ b3 is a

of two cubes.

3

3

3. The binomial a Ϫ b is a

of two cubes.

4. If a3 Ϫ b3 is divided by a Ϫ b, then the remainder

is

.

True or false?

5. For any real number x, x2 Ϫ 4 ϭ (x Ϫ 2)2.

6. The trinomial 4x2 ϩ 6x ϩ 9 is a perfect square

trinomial.

7. The binomial 4y2 ϩ 25 is prime.

8. If the GCF is not 1, then you should factor it out

first.

9. You can factor y2 Ϫ 5y Ϫ my ϩ 5m by grouping.

10. You can factor x2 ϩ ax Ϫ 3x ϩ 3a by grouping.

Exercises

U Study Tips V

• If you have a choice, sit at the front of the class. It is easier to stay alert when you are at the front.

• If you miss what is going on in class, you miss what your instructor feels is important and most likely to appear on tests and quizzes.

U1V Factoring a Difference or Sum of Two Cubes

Factor each difference or sum of cubes. See Example 1.

1. m3 Ϫ 1

2. z3 Ϫ 27

3.

4.

5.

6.

x3 ϩ 8

y3 ϩ 27

a3 ϩ 125

b3 Ϫ 216

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