3 Factoring the Trinomial ax[Sup(2)] + bx + c with a = 1
Tải bản đầy đủ - 0trang
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The only integers that have a product of 6 and a sum of 5 are 2 and 3. Now replace
5x with 2x ϩ 3x and factor by grouping:
x 2 ϩ 5x ϩ 6 ϭ x 2 ϩ 2x ϩ 3x ϩ 6
Replace 5x by 2x ϩ 3x.
ϭ x(x ϩ 2) ϩ 3(x ϩ 2)
Factor out x and 3.
ϭ (x ϩ 3)(x ϩ 2)
Factor out x ϩ 2.
Check by FOIL: (x ϩ 3)(x ϩ 2) ϭ x2 ϩ 5x ϩ 6.
b) To factor x 2 ϩ 8x ϩ 12, we need two integers that have a product of 12 and a sum
of 8. Since the product and sum are both positive, both integers are positive.
Product
Sum
12 ϭ 1 и 12
12 ϭ 2 и 6
12 ϭ 3 и 4
1 ϩ 12 ϭ 13
2ϩ6ϭ8
3ϩ4ϭ7
The only integers that have a product of 12 and a sum of 8 are 2 and 6. Now
replace 8x by 2x ϩ 6x and factor by grouping:
x 2 ϩ 8x ϩ 12 ϭ x 2 ϩ 2x ϩ 6x ϩ 12
Replace 8x by 2x ϩ 6x.
ϭ x(x ϩ 2) ϩ 6(x ϩ 2) Factor out x and 6.
ϭ (x ϩ 6)(x ϩ 2)
Factor out x ϩ 2.
Check by FOIL: (x ϩ 6)(x ϩ 2) ϭ x 2 ϩ 8x ϩ 12.
c) To factor a2 Ϫ 9a ϩ 20, we need two integers that have a product of 20 and a sum
of Ϫ9. Since the product is positive and the sum is negative, both integers must be
negative.
Product
Sum
20 ϭ (Ϫ1)(Ϫ20)
20 ϭ (Ϫ2)(Ϫ10)
20 ϭ (Ϫ4)(Ϫ5)
Ϫ1 ϩ (Ϫ20) ϭ Ϫ21
Ϫ2 ϩ (Ϫ10) ϭ Ϫ12
Ϫ4 ϩ (Ϫ5) ϭ Ϫ9
Only Ϫ4 and Ϫ5 have a product of 20 and a sum of Ϫ9. Now replace Ϫ9a by
Ϫ4a ϩ (Ϫ5a) or Ϫ4a Ϫ 5a and factor by grouping:
a2 Ϫ 9a ϩ 20 ϭ a2 Ϫ 4a Ϫ 5a ϩ 20
Replace Ϫ9a by Ϫ4a Ϫ 5a.
ϭ a(a Ϫ 4) Ϫ 5(a Ϫ 4) Factor out a and Ϫ5.
ϭ (a Ϫ 5)(a Ϫ 4)
Factor out a Ϫ 4.
Check by FOIL: (a Ϫ 5)(a Ϫ 4) ϭ a2 Ϫ 9a ϩ 20.
Now do Exercises 1–14
We usually do not write out all of the steps shown in Example 1. We saw prior to
Example 1 that
x2 ϩ (m ϩ n)x ϩ mn ϭ (x ϩ m)(x ϩ n).
So once you know m and n, you can simply write the factors, as shown in Example 2.
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E X A M P L E
5.3
2
Factoring the Trinomial ax2 ϩ bx ϩ c with a ϭ 1
341
Factoring trinomials more efficiently
Factor.
a) x 2 ϩ 5x ϩ 4
b) y2 ϩ 6y Ϫ 16
c) w2 Ϫ 5w Ϫ 24
Solution
a) To factor x2 ϩ 5x ϩ 4 we need two integers with a product of 4 and a sum of 5.
The only possibilities for a product of 4 are
(1)(4), (Ϫ1)(Ϫ4), (2)(2), and (Ϫ2)(Ϫ2).
Only 1 and 4 have a sum of 5. So,
x 2 ϩ 5x ϩ 4 ϭ (x ϩ 1)(x ϩ 4).
Check by using FOIL on (x ϩ 1)(x ϩ 4) to get x2 ϩ 5x ϩ 4.
b) To factor y 2 ϩ 6y Ϫ 16 we need two integers with a product of Ϫ16 and a sum of 6.
The only possibilities for a product of Ϫ16 are
(Ϫ1)(16), (1)(Ϫ16), (Ϫ2)(8), (2)(Ϫ8), and (Ϫ4)(4).
Only Ϫ2 and 8 have a sum of 6. So,
y 2 ϩ 6y Ϫ 16 ϭ (y ϩ 8)( y Ϫ 2).
Check by using FOIL on ( y ϩ 8)( y Ϫ 2) to get y 2 ϩ 6y Ϫ 16.
c) To factor w2 Ϫ 5w Ϫ 24 we need two integers with a product of Ϫ24 and a sum
of Ϫ5. The only possibilities for a product of Ϫ24 are
(Ϫ1)(24), (1)(Ϫ24), (Ϫ2)(12), (2)(Ϫ12), (Ϫ3)(8), (3)(Ϫ8), (Ϫ4)(6), and (4)(Ϫ6).
Only Ϫ8 and 3 have a sum of Ϫ5. So,
w2 Ϫ 5w Ϫ 24 ϭ (w Ϫ 8)(w ϩ 3).
Check by using FOIL on (w Ϫ 8)(w ϩ 3) to get w2 Ϫ 5w Ϫ 24.
Now do Exercises 15–22
Polynomials are easiest to factor when they are in the form ax 2 ϩ bx ϩ c. So if a
polynomial can be rewritten into that form, rewrite it before attempting to factor it. In
Example 3, we factor polynomials that need to be rewritten.
E X A M P L E
3
Factoring trinomials
Factor.
a) 2x Ϫ 8 ϩ x2
b) Ϫ36 ϩ t 2 Ϫ 9t
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Solution
a) Before factoring, write the trinomial as x 2 ϩ 2x Ϫ 8. Now, to get a product of Ϫ8
and a sum of 2, use Ϫ2 and 4:
2x Ϫ 8 ϩ x 2 ϭ x 2 ϩ 2x Ϫ 8
Write in ax2 ϩ bx ϩ c form.
ϭ (x ϩ 4)(x Ϫ 2) Factor and check by multiplying.
b) Before factoring, write the trinomial as t 2 Ϫ 9t Ϫ 36. Now, to get a product of Ϫ36
and a sum of Ϫ9, use Ϫ12 and 3:
Ϫ36 ϩ t 2 Ϫ 9t ϭ t 2 Ϫ 9t Ϫ 36
Write in ax2 ϩ bx ϩ c form.
ϭ (t Ϫ 12)(t ϩ 3) Factor and check by multiplying.
Now do Exercises 23–24
To factor x 2 ϩ bx ϩ c, we search through all pairs of integers that have a product
of c until we find a pair that has a sum of b. If there is no such pair of integers, then
the polynomial cannot be factored and it is a prime polynomial. Before you can conclude that a polynomial is prime, be sure that you have tried all possibilities.
E X A M P L E
4
Prime polynomials
Factor.
a) x 2 ϩ 7x Ϫ 6
b) x 2 ϩ 9
Solution
a) Because the last term is Ϫ6, we want a positive integer and a negative integer
that have a product of Ϫ6 and a sum of 7. Check all possible pairs of integers:
Product
Sum
Ϫ6 ϭ (Ϫ1)(6)
Ϫ1 ϩ 6 ϭ 5
Ϫ6 ϭ (1)(Ϫ6)
1 ϩ (Ϫ6) ϭ Ϫ5
Ϫ6 ϭ (2)(Ϫ3)
2 ϩ (Ϫ3) ϭ Ϫ1
Ϫ6 ϭ (Ϫ2)(3)
U Helpful Hint V
Don’t confuse a2 ϩ b2 with the difference of two squares a2 Ϫ b2 which is
not a prime polynomial:
a2 Ϫ b2 ϭ (a ϩ b)(a Ϫ b)
Ϫ2 ϩ 3 ϭ 1
None of these possible factors of Ϫ6 have a sum of 7, so we can be certain that
x2 ϩ 7x Ϫ 6 cannot be factored. It is a prime polynomial.
b) Because the x-term is missing in x2 ϩ 9, its coefficient is 0. That is, x2 ϩ 9 ϭ
x2 ϩ 0x ϩ 9. So we seek two positive integers or two negative integers that have
a product of 9 and a sum of 0. Check all possibilities:
Product
9 ϭ (1)(9)
9 ϭ (Ϫ1)(Ϫ9)
9 ϭ (3)(3)
9 ϭ (Ϫ3)(Ϫ3)
Sum
1 ϩ 9 ϭ 10
Ϫ1 ϩ (Ϫ9) ϭ Ϫ10
3ϩ3ϭ6
Ϫ3 ϩ (Ϫ3) ϭ Ϫ6
None of these pairs of integers have a sum of 0, so we can conclude that x 2 ϩ 9 is
a prime polynomial. Note that x 2 ϩ 9 does not factor as (x ϩ 3)2 because
(x ϩ 3)2 has a middle term: (x ϩ 3)2 ϭ x2 ϩ 6x ϩ 9.
Now do Exercises 25–52
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5.3
343
The prime polynomial x 2 ϩ 9 in Example 4(b) is a sum of two squares. There are
many other sums of squares that are prime. For example,
x2 ϩ 1,
a2 ϩ 4,
b2 ϩ 9, and 4y2 ϩ 25
are prime. However, not every sum of two squares is prime. For example, 4x2 ϩ 16 is
a sum of two squares that is not prime because 4x2 ϩ 16 ϭ 4(x2 ϩ 4).
Sum of Two Squares
The sum of two squares a2 ϩ b2 is prime, but not every sum of two squares is prime.
U2V Factoring with Two Variables
In Example 5, we factor polynomials that have two variables using the same technique
that we used for one variable.
E X A M P L E
5
Polynomials with two variables
Factor.
a) x 2 ϩ 2xy Ϫ 8y 2
b) a 2 Ϫ 7ab ϩ 10b2
c) 1 Ϫ 2xy Ϫ 8x2y2
Solution
a) To factor x2 ϩ 2xy Ϫ 8y2 we need two integers with a product of Ϫ8 and a sum
of 2. The only possibilities for a product of Ϫ8 are
(Ϫ1)(8), (1)(Ϫ8), (Ϫ2)(4), and (2)(Ϫ4).
Only Ϫ2 and 4 have a sum of 2. Since (Ϫ2y)(4y) ϭ Ϫ8y 2, we have
x 2 ϩ 2xy Ϫ 8y 2 ϭ (x Ϫ 2y)(x ϩ 4y).
Check by using FOIL on (x Ϫ 2y)(x ϩ 4y) to get x2 ϩ 2xy Ϫ 8y2.
b) To factor a2 Ϫ 7ab ϩ 10b2 we need two integers with a product of 10 and a sum
of Ϫ7. The only possibilities for a product of 10 are
(Ϫ1)(Ϫ10), (1)(10), (Ϫ2)(Ϫ5), and (2)(5).
Only Ϫ2 and Ϫ5 have a sum of Ϫ7. Since (Ϫ2b)(Ϫ5b) ϭ 10b2, we have
a2 Ϫ 7ab ϩ 10b2 ϭ (a Ϫ 5b)(a Ϫ 2b).
Check by using FOIL on (a Ϫ 2b)(a Ϫ 5b) to get a2 Ϫ 7ab ϩ 10b2.
c) As in part (a), we need two integers with a product of Ϫ8 and a sum of Ϫ2. The
integers are Ϫ4 and 2. Since 1 factors as 1 и 1 and Ϫ8x2y2 ϭ (Ϫ4xy)(2xy), we have
1 Ϫ 2xy Ϫ 8x2y2 ϭ (1 ϩ 2xy)(1 Ϫ 4xy).
Check by using FOIL.
Now do Exercises 53–64
U3V Factoring Completely
In Section 5.2 you learned that binomials such as 3x Ϫ 5 (with no common factor) are
prime polynomials. In Example 4 of this section we saw a trinomial that is a prime
polynomial. There are infinitely many prime trinomials. When factoring a polynomial
completely, we could have a factor that is a prime trinomial.
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Chapter 5 Factoring
E X A M P L E
6
Factoring completely
Factor each polynomial completely.
a) x3 Ϫ 6x2 Ϫ 16x
b) 4x3 ϩ 4x2 ϩ 4x
Solution
a) x3 Ϫ 6x 2 Ϫ 16x ϭ x (x 2 Ϫ 6x Ϫ 16) Factor out the GCF.
ϭ x(x Ϫ 8)(x ϩ 2)
Factor x2 Ϫ 6x Ϫ 16.
b) First factor out 4x, the greatest common factor:
4x3 ϩ 4x2 ϩ 4x ϭ 4x (x2 ϩ x ϩ 1)
To factor x2 ϩ x ϩ 1, we would need two integers with a product of 1 and a sum
of 1. Because there are no such integers, x2 ϩ x ϩ 1 is prime, and the factorization
is complete.
Now do Exercises 65–106
Warm-Ups
▼
Fill in the blank.
5.3
1. If there are no two integers that have a
of c and
2
a
of b, then x ϩ bx ϩ c is prime.
2. We can check all factoring by
the factors.
3. The sum of two squares a2 ϩ b2 is
.
4. Always factor out the
first.
True or false?
5.
6.
7.
8.
9.
10.
11.
x2 Ϫ 6x ϩ 9 ϭ (x Ϫ 3)2
x2 ϩ 6x ϩ 9 ϭ (x ϩ 3)2
x2 ϩ 10x ϩ 9 ϭ (x Ϫ 9)(x Ϫ 1)
x2 ϩ 8x Ϫ 9 ϭ (x Ϫ 1)(x ϩ 9)
x2 Ϫ 10xy ϩ 9y2 ϭ (x Ϫ y)(x Ϫ 9y)
x2 ϩ 1 ϭ (x ϩ 1)(x ϩ 1)
x2 ϩ x ϩ1 ϭ (x ϩ 1)(x ϩ 1)
Exercises
U Study Tips V
• Put important facts on note cards. Work on memorizing the note cards when you have a few spare minutes.
• Post some note cards on your refrigerator door. Make this course a part of your life.
U1V Factoring ax2 ؉ bx ؉ c with a ؍1
Factor each trinomial. Write out all of the steps as shown in
Example 1. See the Strategy for Factoring x2 ϩ bx ϩ c by
Grouping on page 339.
1. x 2 ϩ 4x ϩ 3
2. y 2 ϩ 6y ϩ 5
3. x 2 ϩ 9x ϩ 18
4. w 2 ϩ 6w ϩ 8
5. a2 ϩ 7a ϩ 10
6. b2 ϩ 7b ϩ 12
7. a2 Ϫ 7a ϩ 12
8. m 2 Ϫ 9m ϩ 14
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5.3
9. b2 Ϫ 5b Ϫ 6
10. a 2 ϩ 5a Ϫ 6
11. x2 ϩ 3x Ϫ 10
12. x2 Ϫ x Ϫ 12
13. x2 ϩ 5x Ϫ 24
14. a2 Ϫ 5a Ϫ 50
Factoring the Trinomial ax2 ϩ bx ϩ c with a ϭ 1
345
49. x2 Ϫ 5x Ϫ 150
50. x2 Ϫ 25x ϩ 150
51. 13y ϩ 30 ϩ y2
52. 18z ϩ 45 ϩ z2
U2V Factoring with Two Variables
Factor each polynomial. If the polynomial is prime, say so.
See Examples 2– 4.
Factor each polynomial. See Example 5.
15. y 2 ϩ 7y ϩ 10
54. a2 ϩ 7ab ϩ 10b2
16. x 2 ϩ 8x ϩ 15
55. x2 Ϫ 4xy Ϫ 12y 2
17. a 2 Ϫ 6a ϩ 8
56. y 2 ϩ yt Ϫ 12t 2
18. b2 Ϫ 8b ϩ 15
57. x 2 Ϫ 13xy ϩ 12y2
19. m Ϫ 10m ϩ 16
58. h2 Ϫ 9hs ϩ 9s 2
20. m 2 Ϫ 17m ϩ 16
59. x 2 ϩ 4xz Ϫ 33z2
21. w 2 ϩ 9w Ϫ 10
60. x 2 Ϫ 5xs Ϫ 24s2
22. m ϩ 6m Ϫ 16
61. 1 ϩ 3ab Ϫ 28a2b2
23. w Ϫ 8 Ϫ 2w
62. 1 Ϫ xy Ϫ 20x2y2
24. Ϫ16 ϩ m 2 Ϫ 6m
63. 15a2b2 ϩ 8ab ϩ 1
25. a 2 Ϫ 2a Ϫ 12
64. 12m2n2 Ϫ 8mn ϩ 1
2
2
2
53. x2 ϩ 5ax ϩ 6a2
26. x ϩ 3x ϩ 3
2
27. 15m Ϫ 16 ϩ m2
28. 3y ϩ y Ϫ 10
2
29. a 2 Ϫ 4a ϩ 12
30. y 2 Ϫ 6y Ϫ 8
31. z 2 Ϫ 25
32. p2 Ϫ 1
33. h2 ϩ 49
34. q2 ϩ 4
U3V Factoring Completely
Factor each polynomial completely. Use the methods discussed
in Sections 5.1 through 5.3. If the polynomial is prime say so.
See Example 6.
65. 5x3 ϩ 5x
66. b3 ϩ 49b
67. w2 Ϫ 8w
68. x4 Ϫ x3
35. m2 ϩ 12m ϩ 20
69. 2w 2 Ϫ 162
36. m2 ϩ 21m ϩ 20
70. 6w4 Ϫ 54w2
37. t2 Ϫ 3t ϩ 10
71. Ϫ2b2 Ϫ 98
38. x2 Ϫ 5x Ϫ 3
72. Ϫa3 Ϫ 100a
39. m2 Ϫ 18 Ϫ 17m
73. x3 Ϫ 2x2 Ϫ 9x ϩ 18
40. h2 Ϫ 36 ϩ 5h
74. x3 ϩ 7x2 Ϫ x Ϫ 7
41. m2 Ϫ 23m ϩ 24
75. 4r2 ϩ 9
42. m2 ϩ 23m ϩ 24
76. t2 ϩ 4z2
43. 5t Ϫ 24 ϩ t 2
77. x 2w 2 ϩ 9x2
44. t2 Ϫ 24 Ϫ 10t
78. a4b ϩ a2b3
45. t2 Ϫ 2t Ϫ 24
79. w2 Ϫ 18w ϩ 81
46. t2 ϩ 14t ϩ 24
80. w2 ϩ 30w ϩ 81
47. t2 Ϫ 10t Ϫ 200
81. 6w2 Ϫ 12w Ϫ 18
48. t2 ϩ 30t ϩ 200
82. 9w Ϫ w3
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83. 3y2 ϩ 75
108. Area of a sail. The area in square meters for a triangular
sail is given by A(x) ϭ x2 ϩ 5x ϩ 6.
84. 5x2 ϩ 500
a) Find A(5).
b) If the height of the sail is x ϩ 3 meters, then what is
the length of the base of the sail?
85. ax ϩ ay ϩ cx ϩ cy
86. y3 ϩ y2 Ϫ 4y Ϫ 4
87. Ϫ2x2 Ϫ 10x Ϫ 12
88. Ϫa3 Ϫ 2a2 Ϫ a
89. 32x2 Ϫ 2x4
90. 20w 2 ϩ 100w ϩ 40
91. 3w2 ϩ 27w ϩ 54
xϩ3m
92. w3 Ϫ 3w2 Ϫ 18w
93. 18w2 ϩ w3 ϩ 36w
94. 18a2 ϩ 3a3 ϩ 36a
Base
Area ϭ x 2 ϩ 5x ϩ 6 m 2
95. 9y2 ϩ 1 ϩ 6y
96. 2a2 ϩ 1 ϩ 3a
Figure for Exercise 108
97. 8vw2 ϩ 32vw ϩ 32v
98. 3h2t ϩ 6ht ϩ 3t
109. Volume of a cube. Hector designed a cubic box with
volume x 3 cubic feet. After increasing the dimensions
of the bottom, the box has a volume of x 3 ϩ 8x 2 ϩ 15x
cubic feet. If each of the dimensions of the bottom
was increased by a whole number of feet, then how
much was each increase?
99. 6x 3y ϩ 30x 2 y 2 ϩ 36xy3
100. 3x 3y 2 Ϫ 3x 2y 2 ϩ 3xy 2
101. 5 ϩ 8w ϩ 3w2
102. Ϫ3 ϩ 2y ϩ 21y2
103. Ϫ3y3 ϩ 6y2 Ϫ 3y
104. Ϫ4w3 Ϫ 16w2 ϩ 20w
105. a3 ϩ ab ϩ 3b ϩ 3a2
106. ac ϩ xc ϩ aw2 ϩ xw2
Applications
Use factoring to solve each problem.
107. Area of a deck. The area in square feet for a rectangular
deck is given by A(x) ϭ x 2 ϩ 6x ϩ 8.
a) Find A(6).
b) If the width of the deck is x ϩ 2 feet, then what is the
length?
110. Volume of a container. A cubic shipping container
had a volume of a3 cubic meters. The height was
decreased by a whole number of meters and the
width was increased by a whole number of meters so
that the volume of the container is now a3 ϩ 2a2 Ϫ 3a
cubic meters. By how many meters were the height
and width changed?
Getting More Involved
111. Discussion
Which of the following products is not equivalent
to the others? Explain your answer.
a) (2x Ϫ 4)(x ϩ 3)
c) 2(x Ϫ 2)(x ϩ 3)
b) (x Ϫ 2)(2x ϩ 6)
d) (2x Ϫ 4)(2x ϩ 6)
112. Discussion
L
Area ϭ x 2 ϩ 6x ϩ 8 ft 2
Figure for Exercise 107
x ϩ 2 ft
When asked to factor completely a certain polynomial,
four students gave the following answers. Only one
student gave the correct answer. Which one must it be?
Explain your answer.
a) 3(x 2 Ϫ 2x Ϫ 15)
c) 3(x Ϫ 5)(x Ϫ 3)
b) (3x Ϫ 5)(5x Ϫ 15)
d) (3x Ϫ 15)(x Ϫ 3)
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5.4
Mid-Chapter Quiz
Sections 5.1 through 5.3
Find the greatest common factor for each group of integers.
Factor each expression by factoring out the greatest common
factor.
6. 12x Ϫ 30x
3
2
2 2
13. 10x3 Ϫ 250x
14. Ϫ6x2 Ϫ 36x Ϫ 54
15. aw Ϫ 3w ϩ 6a Ϫ 18
7. 15ab Ϫ 25a b ϩ 35a b
3
11. 4h2 ϩ 12h ϩ 9
12. w2 Ϫ 16w ϩ 64
4. 60, 144, 240
5. 8w Ϫ 6y
Chapter 5
10. 4y2 Ϫ 9w2
2. 140
3. 36, 45
347
1
Factor completely.
Find the prime factorization of each integer.
1. 48
Factoring the Trinomial ax2 ϩ bx ϩ c with a
3
16. bx Ϫ 5b Ϫ 6x ϩ 30
Factor each expression.
17. ax2 Ϫ a ϩ x2 Ϫ 1
8. (x ϩ 3)x Ϫ (x ϩ 3)5
18. x3 Ϫ 5x Ϫ 4x2
9. m(m Ϫ 9) Ϫ 6(m Ϫ 9)
19. 2x3 ϩ 18x
20. a2 Ϫ 12as ϩ 32s2
5.4
In This Section
U1V The ac Method
U2V Trial and Error
U3V Factoring Completely
Factoring the Trinomial ax2 ؉ bx ؉ c with a
1
In Section 5.3, we used grouping to factor trinomials with a leading coefficient of 1.
In this section we will also use grouping to factor trinomials with a leading
coefficient that is not equal to 1.
U1V The ac Method
The first step in factoring ax2 ϩ bx ϩ c with a ϭ 1 is to find two numbers with a product of c and a sum of b. If a 1, then the first step is to find two numbers with a
product of ac and a sum of b. This method is called the ac method. The strategy for
factoring by the ac method follows. Note that this strategy works whether or not the
leading coefficient is 1.
Strategy for Factoring ax 2 ؉ bx ؉ c by the ac Method
To factor the trinomial ax2 ϩ bx ϩ c:
1. Find two numbers that have a product equal to ac and a sum equal to b.
2. Replace bx by the sum of two terms whose coefficients are the two numbers
found in (1).
3. Factor the resulting four-term polynomial by grouping.
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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