5: Factoring, Solving Equations, and Problem Solving
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256
Chapter 6 • Factoring, Solving Equations, and Problem Solving
Likewise, 25x2 Ϫ 40xy ϩ 16y2 is a perfect square trinomial for these reasons:
1. The first term is a square: (5x) 2.
2. The last term is a square: (4y) 2.
3. The middle term is twice the product of the quantities being squared in the first and
last terms: 2(5x)(4y).
Once we know that we have a perfect square trinomial, the factoring process follows immediately from the two basic patterns.
9x2 ϩ 30x ϩ 25 ϭ (3x ϩ 5) 2
25x2 Ϫ 40xy ϩ 16y2 ϭ (5x Ϫ 4y) 2
Here are some additional examples of perfect square trinomials and their factored
form.
x2 Ϫ 16x ϩ 64 ϭ
16x2 Ϫ 56x ϩ 49 ϭ
25x2 ϩ 20xy ϩ 4y2 ϭ
1 ϩ 6y ϩ 9y2 ϭ
4m2 Ϫ 4mn ϩ n2 ϭ
(x) 2 Ϫ 2(x) (8) ϩ (8) 2
(4x) 2 Ϫ 2(4x) (7) ϩ (7) 2
(5x) 2 ϩ 2(5x) (2y) ϩ (2y) 2
(1) 2 ϩ 2(1) (3y) ϩ (3y) 2
(2m) 2 Ϫ 2(2m)(n) ϩ (n) 2
ϭ
ϭ
ϭ
ϭ
ϭ
(x Ϫ 8) 2
(4x Ϫ 7) 2
(5x ϩ 2y) 2
(1 ϩ 3y) 2
(2m Ϫ n) 2
You may want to do this step mentally after
you feel comfortable with the process
We have considered some basic factoring techniques in this chapter one at a time, but you
must be able to apply them as needed in a variety of situations. So, let’s first summarize the
techniques and then consider some examples.
In this chapter we have discussed these techniques:
1. Factoring by using the distributive property to factor out the greatest common monomial or binomial factor
2. Factoring by grouping
3. Factoring by applying the difference-of-squares pattern
4. Factoring by applying the perfect-square-trinomial pattern
5. Factoring trinomials of the form x2 ϩ bx ϩ c into the product of two binomials
6. Factoring trinomials of the form ax2 ϩ bx ϩ c into the product of two binomials
As a general guideline, always look for a greatest common monomial factor first, and then
proceed with the other factoring techniques.
In each of the following examples we have factored completely whenever possible. Study
them carefully, and notice the factoring techniques we used.
1. 2x2 ϩ 12x ϩ 10 ϭ 2(x2 ϩ 6x ϩ 5) ϭ 2(x ϩ 1)(x ϩ 5)
2. 4x2 ϩ 36 ϭ 4(x2 ϩ 9)
Remember that the sum of two squares is not factorable
using integers unless there is a common factor
3. 4t2 ϩ 20t ϩ 25 ϭ (2t ϩ 5) 2
If you fail to recognize a perfect trinomial square, no
harm is done; simply proceed to factor into the product
of two binomials, and then you will recognize that the
two binomials are the same
6.5 • Factoring, Solving Equations, and Problem Solving
257
4. x2 Ϫ 3x Ϫ 8 is not factorable using integers. This becomes obvious from the table.
Product
1(Ϫ8)
Ϫ1(8)
2(Ϫ4)
Ϫ2(4)
Sum
ϭ Ϫ8 1 ϩ (Ϫ8) ϭ Ϫ7
ϭ Ϫ8
Ϫ1 ϩ 8 ϭ 7
ϭ Ϫ8 2 ϩ (Ϫ4) ϭ Ϫ2
ϭ Ϫ8
Ϫ2 ϩ 4 ϭ 2
No two factors of Ϫ8 produce a sum of Ϫ3.
5. 6y2 Ϫ 13y Ϫ 28 ϭ (2y Ϫ 7)(3y ϩ 4) . We found the binomial factors as follows:
(y ϩ _____)(6y Ϫ _____)
or
(y Ϫ _____)(6y ϩ _____)
or
(2y Ϫ _____)(3y ϩ _____)
or
(2y ϩ _____)(3y Ϫ _____)
Factors of 28
1 и 28
2 и 14
4и7
or
or
or
28 и 1
14 и 2
7и4
6. 32x2 Ϫ 50y2 ϭ 2(16x2 Ϫ 25y2 ) ϭ 2(4x ϩ 5y)(4x Ϫ 5y)
Solving Equations by Factoring
As stated in the preface, there is a common thread that runs throughout this text: namely, learn
a skill; next, use the skill to help solve equations; and then use equations to help solve application problems. This thread becomes evident in this chapter. After presenting a factoring
technique, we immediately solved some equations using this technique, and then considered
some applications involving such equations. The following steps summarize the equation
solving process in this chapter.
1. Organize all terms of the polynomial on the same side of the equation with zero on
the other side.
2. Factor the polynomial. This will involve a variety of factoring techniques presented
in this chapter.
3. Set each factor equal to zero and solve for the unknown.
4. Check your solutions back into the original equation.
Let’s consider some examples.
Classroom Example
Solve x2 ϭ 64x.
EXAMPLE 1
Solve x2 ϭ 25x.
Solution
x2 ϭ 25x
x Ϫ 25x ϭ 0
Added Ϫ25x to both sides
x(x Ϫ 25) ϭ 0
x ϭ 0 or x Ϫ 25 ϭ 0
x ϭ 25
x ϭ 0 or
2
The solution set is {0, 25}. Check it!
258
Chapter 6 • Factoring, Solving Equations, and Problem Solving
Classroom Example
Solve m3 Ϫ 144m ϭ 0.
Solve x3 Ϫ 36x ϭ 0.
EXAMPLE 2
Solution
x3 Ϫ 36x ϭ 0
x(x2 Ϫ 36) ϭ 0
x(x ϩ 6)(x Ϫ 6) ϭ 0
x ϭ 0 or x ϩ 6 ϭ 0 or x Ϫ 6 ϭ 0 If abc ϭ 0, then a ϭ 0 or b ϭ 0 or c ϭ 0
xϭ6
x ϭ 0 or x ϭ Ϫ6 or
The solution set is {Ϫ6, 0, 6}. Does it check?
Classroom Example
Solve 8x2 ϩ 10x Ϫ 7 ϭ 0.
Solve 10x2 Ϫ 13x Ϫ 3 ϭ 0.
EXAMPLE 3
Solution
10x2 Ϫ 13x Ϫ 3 ϭ 0
(5x ϩ 1)(2x Ϫ 3) ϭ 0
5x ϩ 1 ϭ 0 or 2x Ϫ 3 ϭ 0
5x ϭ Ϫ1 or
2x ϭ 3
1
3
x ϭ Ϫ or
xϭ
5
2
1 3
The solution set is eϪ , f . Does it check?
5 2
Classroom Example
Solve 9x2 Ϫ 24x ϩ 16 ϭ 0.
Solve 4x2 Ϫ 28x ϩ 49 ϭ 0.
EXAMPLE 4
Solution
4x2 Ϫ 28x ϩ 49 ϭ 0
(2x Ϫ 7) 2 ϭ 0
(2x Ϫ 7)(2x Ϫ 7) ϭ 0
2x Ϫ 7 ϭ 0 or 2x Ϫ 7 ϭ 0
2x ϭ 7
2x ϭ 7 or
7
7
x ϭ or
xϭ
2
2
7
The solution set is e f .
2
Pay special attention to the next example. We need to change the form of the original
equation before we can apply the property ab ϭ 0 if and only if a ϭ 0 or b ϭ 0. A necessary
condition of this property is that an indicated product is set equal to zero.
Classroom Example
Solve (x ϩ 6)(x ϩ 3) ϭ 4.
Solve (x ϩ 1) (x ϩ 4) ϭ 40.
EXAMPLE 5
Solution
(x ϩ 1)(x ϩ 4)
x2 ϩ 5x ϩ 4
x2 ϩ 5x Ϫ 36
(x ϩ 9)(x Ϫ 4)
ϭ 40
ϭ 40
ϭ0
ϭ0
6.5 • Factoring, Solving Equations, and Problem Solving
259
x ϩ 9 ϭ 0 or x Ϫ 4 ϭ 0
x ϭ Ϫ9 or
xϭ4
The solution set is {Ϫ9, 4}. Check it!
Classroom Example
Solve 3a2 Ϫ 6a Ϫ 45 ϭ 0.
EXAMPLE 6
Solve 2n2 ϩ 16n Ϫ 40 ϭ 0.
Solution
2n2 ϩ 16n Ϫ 40
2(n2 ϩ 8n Ϫ 20)
n2 ϩ 8n Ϫ 20
(n ϩ 10)(n Ϫ 2)
n ϩ 10 ϭ 0
ϭ0
ϭ0
1
ϭ 0 Multiplied both sides by
2
ϭ0
or n Ϫ 2 ϭ 0
nϭ2
n ϭ Ϫ10 or
The solution set is {Ϫ10, 2}. Does it check?
Problem Solving
Reminder: Throughout this book we highlight the need to learn a skill, to use that skill to
help solve equations, and then to use equations to help solve problems. Our new factoring
skills have provided more ways of solving equations, which in turn gives us more power to
solve word problems. We conclude the chapter by solving a few more problems.
Classroom Example
Find two numbers whose product is
70 if one of the numbers is 1 less
than three times the other number.
EXAMPLE 7
Find two numbers whose product is 65 if one of the numbers is 3 more than twice the other
number.
Solution
Let n represent one of the numbers; then 2n ϩ 3 represents the other number. Since their product is 65, we can set up and solve the following equation:
n(2n ϩ 3) ϭ 65
2n ϩ 3n Ϫ 65 ϭ 0
(2n ϩ 13)(n Ϫ 5) ϭ 0
2n ϩ 13 ϭ 0 or n Ϫ 5 ϭ 0
2n ϭ Ϫ13 or
nϭ5
13
n ϭ Ϫ or
nϭ5
2
2
13
13
, then 2n ϩ 3 ϭ 2aϪ b ϩ 3 ϭ Ϫ10. If n ϭ 5, then 2n ϩ 3 ϭ 2(5) ϩ 3
2
2
13
ϭ 13. Thus the numbers are Ϫ and Ϫ10, or 5 and 13.
2
If n ϭ Ϫ
Classroom Example
The area of a rectangular poster is
48 square centimeters. The length is
8 centimeters longer than the width
of the poster. Find the length and
width of the rectangular poster.
EXAMPLE 8
The area of a triangular sheet of paper is 14 square inches. One side of the triangle is 3 inches longer than the altitude to that side. Find the length of the one side and the length of the
altitude to that side.
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Chapter 6 • Factoring, Solving Equations, and Problem Solving
Solution
Let h represent the altitude to the side. Then h ϩ 3 represents the side of the triangle (see
Figure 6.5).
h
h+3
Figure 6.5
1
Since the formula for finding the area of a triangle is A ϭ bh, we have
2
1
h(h ϩ 3) ϭ 14
2
h(h ϩ 3) ϭ 28 Multiplied both sides by 2
h2 ϩ 3h ϭ 28
h2 ϩ 3h Ϫ 28 ϭ 0
(h ϩ 7)(h Ϫ 4) ϭ 0
h ϩ 7 ϭ 0 or h Ϫ 4 ϭ 0
hϭ4
h ϭ Ϫ7 or
The solution of Ϫ7 is not reasonable. Thus the altitude is 4 inches and the length of the side
to which that altitude is drawn is 7 inches.
Classroom Example
A photograph measures 14 inches
wide and 19 inches long. A strip of
uniform width is to be cut off from
both ends and both sides of the
photograph in order to reduce the
area of the photograph to 176 square
inches. Find the width of the strip.
EXAMPLE 9
A strip with a uniform width is shaded along both sides and both ends of a rectangular poster
with dimensions 12 inches by 16 inches. How wide is the strip if one-half of the area of the
poster is shaded?
Solution
Let x represent the width of the shaded strip of the poster in Figure 6.6. The area of the strip
1
is one-half of the area of the poster; therefore, it is (12)(16) ϭ 96 square inches. Furthermore,
2
we can represent the area of the strip around the poster by the words the area of the poster
minus the area of the unshaded portion.
x
x
H
MAT 16 − 2x
N
ART OSITIO12 − 2x
EXP
2010
x
16 inches
Figure 6.6
x
12 inches
6.5 • Factoring, Solving Equations, and Problem Solving
261
Thus we can set up and solve the following equation:
Area of poster Ϫ Area of unshaded portion ϭ Area of strip
16(12)
Ϫ
(16 Ϫ 2x)(12 Ϫ 2x)
ϭ
96
192 Ϫ (192 Ϫ 56x ϩ 4x 2 ) ϭ 96
192 Ϫ 192 ϩ 56x Ϫ 4x 2 ϭ 96
Ϫ4x 2 ϩ 56x Ϫ 96 ϭ 0
x 2 Ϫ 14x ϩ 24 ϭ 0
(x Ϫ 12)(x Ϫ 2) ϭ 0
x Ϫ 12 ϭ 0 or x Ϫ 2 ϭ 0
x ϭ 12 or
x ϭ2
Obviously, the strip cannot be 12 inches wide because the total width of the poster is 12 inches.
Thus we must disregard the solution of 12 and conclude that the strip is 2 inches wide.
Concept Quiz 6.5
For Problems 1–7, match each factoring problem with the name of the type of pattern that
would be used to factor the problem.
1.
2.
3.
4.
5.
6.
7.
x2 ϩ 2xy ϩ y2
x2 Ϫ y2
ax ϩ ay ϩ bx ϩ by
x2 ϩ bx ϩ c
ax2 ϩ bx ϩ c
ax2 ϩ ax ϩ a
(a ϩ b)x ϩ (a ϩ b)y
A.
B.
C.
D.
E.
F.
G.
Trinomial with an x-squared coeffcient of one
Common binomial factor
Difference of two squares
Common factor
Factor by grouping
Perfect-square trinomial
Trinomial with an x-squared coefficient of not one
Problem Set 6.5
For Problems 1–12, factor each of the perfect square trinomials. (Objective 1)
1. x 2 ϩ 4x ϩ 4
2. x 2 ϩ 18x ϩ 81
3. x 2 Ϫ 10x ϩ 25
4. x 2 Ϫ 24x ϩ 144
5. 9n ϩ 12n ϩ 4
6. 25n ϩ 30n ϩ 9
7. 16a Ϫ 8a ϩ 1
8. 36a2 Ϫ 84a ϩ 49
2
2
2
9. 4 ϩ 36x ϩ 81x 2
10. 1 Ϫ 4x ϩ 4x 2
11. 16x 2 Ϫ 24xy ϩ 9y 2
12. 64x ϩ 16xy ϩ y
2
19. 3a2 Ϫ 7a Ϫ 4
20. a2 ϩ 7a Ϫ 30
21. 8x 2 ϩ 72
22. 3y 3 Ϫ 36y 2 ϩ 96y
23. 9x 2 ϩ 30x ϩ 25
24. 5x 2 Ϫ 5x Ϫ 6
25. 15x 2 ϩ 65x ϩ 70
27. 24x 2 ϩ 2x Ϫ 15
26. 4x 2 Ϫ 20xy ϩ 25y 2
28. 9x 2y Ϫ 27xy
29. xy ϩ 5y Ϫ 8x Ϫ 40
30. xy Ϫ 3y ϩ 9x Ϫ 27
31. 20x 2 ϩ 31xy Ϫ 7y 2
2
For Problems 13– 40, factor each polynomial completely.
Indicate any that are not factorable using integers. (Objective 2)
32. 2x 2 Ϫ xy Ϫ 36y 2
33. 24x 2 ϩ 18x Ϫ 81
34. 30x 2 ϩ 55x Ϫ 50
35. 12x 2 ϩ 6x ϩ 30
36. 24x 2 Ϫ 8x ϩ 32
37. 5x 4 Ϫ 80
13. 2x 2 ϩ 17x ϩ 8
15. 2x 3 Ϫ 72x
14. x 2 ϩ 19x
16. 30x 2 Ϫ x Ϫ 1
38. 3x 5 Ϫ 3x
17. n2 Ϫ 7n Ϫ 60
18. 4n3 Ϫ 100n
40. 4x 2 Ϫ 28xy ϩ 49y 2
39. x 2 ϩ 12xy ϩ 36y 2
262
Chapter 6 • Factoring, Solving Equations, and Problem Solving
For Problems 41– 70, solve each equation. (Objective 3)
41. 4x Ϫ 20x ϭ 0
42. Ϫ3x Ϫ 24x ϭ 0
43. x 2 Ϫ 9x Ϫ 36 ϭ 0
44. x 2 ϩ 8x Ϫ 20 ϭ 0
45. Ϫ2x3 ϩ 8x ϭ 0
46. 4x 3 Ϫ 36x ϭ 0
2
2
47. 6n2 Ϫ 29n Ϫ 22 ϭ 0
77. In an office building, a room contains 54 chairs.
The number of chairs per row is three less than twice the
number of rows. Find the number of rows and the number of chairs per row.
78. An apple orchard contains 85 trees. The number of trees
in each row is three less than four times the number of
rows. Find the number of rows and the number of trees
per row.
48. 30n2 Ϫ n Ϫ 1 ϭ 0
49. (3n Ϫ 1)(4n Ϫ 3) ϭ 0
50. (2n Ϫ 3)(7n ϩ 1) ϭ 0
51. (n Ϫ 2)(n ϩ 6) ϭ Ϫ15
52. (n ϩ 3)(n Ϫ 7) ϭ Ϫ25
53. 2x 2 ϭ 12x
54. Ϫ3x2 ϭ 15x
55. t3 Ϫ 2t2 Ϫ 24t ϭ 0
56. 2t3 Ϫ 16t2 Ϫ 18t ϭ 0
57. 12 Ϫ 40x ϩ 25x 2 ϭ 0
58. 12 Ϫ 7x Ϫ 12x 2 ϭ 0
79. Suppose that the combined area of two squares
is 360 square feet. Each side of the larger square is three
times as long as a side of the smaller square. How big is
each square?
80. The area of a rectangular slab of sidewalk is 45 square
feet. Its length is 3 feet more than four times its width.
Find the length and width of the slab.
81. The length of a rectangular sheet of paper is 1 centimeter
more than twice its width, and the area of the rectangle is
55 square centimeters. Find the length and width of the
rectangle.
59. n2 Ϫ 28n ϩ 192 ϭ 0
60. n2 ϩ 33n ϩ 270 ϭ 0
61. (3n ϩ 1)(n ϩ 2) ϭ 12
62. (2n ϩ 5)(n ϩ 4) ϭ Ϫ1
63. x 3 ϭ 6x 2
76. A number is one less than three times another num-ber. If
the product of the two numbers is 102, find the numbers.
64. x3 ϭ Ϫ 4x2
65. 9x 2 Ϫ 24x ϩ 16 ϭ 0
66. 25x 2 ϩ 60x ϩ 36 ϭ 0
67. x 3 ϩ 10x 2 ϩ 25x ϭ 0
68. x 3 Ϫ 18x 2 ϩ 81x ϭ 0
69. 24x 2 ϩ 17x Ϫ 20 ϭ 0
70. 24x 2 ϩ 74x Ϫ 35 ϭ 0
For Problems 71– 88, set up an equation and solve each
problem. (Objective 4)
71. Find two numbers whose product is 15 such that one
of the numbers is seven more than four times the other
number.
72. Find two numbers whose product is 12 such that one of
the numbers is four less than eight times the other
number.
73. Find two numbers whose product is Ϫ1. One of the
numbers is three more than twice the other number.
74. Suppose that the sum of the squares of three consecutive
integers is 110. Find the integers.
75. A number is one more than twice another number. The
sum of the squares of the two numbers is 97. Find the
numbers.
82. Suppose that the length of a certain rectangle is three
times its width. If the length is increased by 2 inches,
and the width increased by 1 inch, the newly formed
rectangle has an area of 70 square inches. Find the
length and width of the original rectangle.
83. The area of a triangle is 51 square inches. One side
of the triangle is 1 inch less than three times the length
of the altitude to that side. Find the length of that side
and the length of the altitude to that side.
84. Suppose that a square and a rectangle have equal areas.
Furthermore, suppose that the length of the rectangle is
twice the length of a side of the square, and the width of
the rectangle is 4 centimeters less than the length of a
side of the square. Find the dimensions of both figures.
85. A strip of uniform width is to be cut off of both sides
and both ends of a sheet of paper that is 8 inches by
11 inches, in order to reduce the size of the paper to an
area of 40 square inches. Find the width of the strip.
86. The sum of the areas of two circles is 100 square centimeters. The length of a radius of the larger circle is
2 centimeters more than the length of a radius of the
smaller circle. Find the length of a radius of each circle.
87. The sum of the areas of two circles is 180 square inches.
The length of a radius of the smaller circle is 6 inches
less than the length of a radius of the larger circle. Find
the length of a radius of each circle.
88. A strip of uniform width is shaded along both sides and
both ends of a rectangular poster that is 18 inches by
14 inches. How wide is the strip if the unshaded portion
of the poster has an area of 165 square inches?
6.5 • Factoring, Solving Equations, and Problem Solving
263
Thoughts Into Words
91. Explain how you would solve (x ϩ 2)(x ϩ 3) ϭ (x ϩ 2)
(3x Ϫ 1). Do you see more than one approach to this
problem?
89. When factoring polynomials, why do you think that it is
best to look for a greatest common monomial factor
first?
90. Explain how you would solve (4x Ϫ 3)(8x ϩ 5) ϭ 0 and
also how you would solve (4x Ϫ 3)(8x ϩ 5) ϭ Ϫ9.
Answers to the Concept Quiz
1. F or A
2. C
3. E
4. A
5. G
6. D
7. B
Chapter 6 Summary
OBJECTIVE
SUMMARY
EXAMPLE
Find the greatest common
factor.
By “the greatest common factor of two or
more monomials” we mean the monomial
with the largest numerical coefficient and
the highest power of the variables, which is
a factor of each given monomial.
Find the greatest common factor of 12a3b3,
18a2b2, and 54ab4.
The distributive property in the form
ab ϩ ac ϭ a(b ϩ c) provides the basis for
factoring out a greatest common monomial
or binomial factor.
Factor 36x2 y ϩ 18x Ϫ 27xy.
(Section 6.1/Objective 1)
Factor out the greatest
common factor.
(Section 6.1/Objective 2)
Factor by grouping.
(Section 6.1/Objective 3)
Factor the difference of two
squares.
(Section 6.2/Objective 1)
Factor trinomials of the form
x2 ϩ bx ϩ c.
(Section 6.3/Objective 1)
Rewriting an expression such as
ab ϩ 3a ϩ bc ϩ 3c as a(b ϩ 3) ϩ c(b ϩ 3)
and then factoring out the common
binomial factor of b ϩ 3 so that
a(b ϩ 3) ϩ c(b ϩ 3) becomes (b ϩ 3)
(a ϩ c), is called factoring by grouping.
The factoring pattern, the difference of two
squares, is a2 Ϫ b2 ϭ (a Ϫ b)(a ϩ b). Be
careful not to apply the pattern to the sum
of two squares, such as a2 ϩ b2. There is
no factoring pattern for the sum of two
squares.
The following multiplication pattern
provides a technique for factoring trinomials
of the form x2 ϩ bx ϩ c.
(x ϩ r)(x ϩ s) ϭ x2 ϩ rx ϩ sx ϩ rs
ϭ x2 ϩ (r ϩ s)x ϩ rs
Solution
The largest numerical coefficient that is a
factor of all three terms is 6. The highest
exponent for a that is a factor of all three
terms is 1. The highest exponent for b that
is a factor of all three terms is 2. So the
greatest common factor is 6ab2.
Solution
36x2 y ϩ 18x Ϫ 27xy
ϭ 9x(4xy) ϩ 9x(2)Ϫ9x(3y)
ϭ 9x(4xy ϩ 2 Ϫ 3y)
Factor 15xy ϩ 6x ϩ 10y2 ϩ 4y.
Solution
15xy ϩ 6x ϩ 10y2 ϩ 4y
ϭ 3x(5y ϩ 2) ϩ 2y(5y ϩ 2)
ϭ (5y ϩ 2)(3x ϩ 2y)
Factor 4x2 Ϫ 81y2.
Solution
4x2 Ϫ 81y2 ϭ (2x ϩ 9y)(2x Ϫ 9y)
Factor x2 ϩ 2x Ϫ 35.
Solution
The factors of Ϫ35 that sum to 2 are 7
and Ϫ5.
x2 ϩ 2x Ϫ 35 ϭ (x ϩ 7)(x Ϫ 5)
For trinomials of the form x2 ϩ bx ϩ c, we
want two factors of c whose sum will be
equal to b.
Factor trinomials in which the
leading coefficient is not 1.
(Section 6.4/Objective 1)
We presented two different techniques
for factoring trinomials of the form
ax2 ϩ bx ϩ c. To review these techniques,
turn to Section 6.4 and study the
examples. The examples here show
the two techniques.
Factor:
(a) 3x2 ϩ 7x ϩ 4
(b) 2x2 Ϫ 5x ϩ 3
Solution
(a) To factor 3x2 ϩ 7x ϩ 4, we can look at
the first term and the sign situation to
(continued)
264
Chapter 6 • Summary
OBJECTIVE
SUMMARY
265
EXAMPLE
determine that the factors are of the form
(3x ϩ
) and (x ϩ
). By trial and
error we can arrive at the correct factors.
3x2 ϩ 7x ϩ 4 ϭ (3x ϩ 4)(x ϩ 1)
(b) 2x2 Ϫ 5x ϩ 3 sum of Ϫ5
product of 2(3) ϭ 6
So we want two numbers whose sum is
Ϫ5 and whose product is 6. The numbers
are Ϫ2 and Ϫ3. Therefore, we rewrite the
middle term as Ϫ2x Ϫ3x.
2x2 Ϫ 5x ϩ 3 ϭ 2x2 Ϫ 2x Ϫ 3x ϩ 3
ϭ 2x(x Ϫ 1) Ϫ 3(x Ϫ 1)
ϭ (x Ϫ 1)(2x Ϫ 3)
Factor perfect-square
trinomials.
(Section 6.5/Objective 1)
Recognize the different types
of factoring.
(Section 6.5/Objective 2)
Perfect-square trinomials are easy to
recognize because of the nature of their
terms. The first term and the last term will
be the squares of a quantity. The middle
term is twice the product of the quantities
being squared in the first and last terms.
Factor 16x2 ϩ 56x ϩ 49.
As a general guideline for factoring
completely, always look for a greatest
common factor first, and then proceed with
one or more of the following techniques.
Factor 3x2 ϩ 12xy ϩ 12y2.
1. Apply the difference-of-squares pattern.
Solution
16x2 ϩ 56x ϩ 49
ϭ (4x2 ) ϩ 2(4x)(7) ϩ (7)2
ϭ (4x ϩ 7)2
Solution
3x2 ϩ 12xy ϩ 12y2
ϭ 3(x2 ϩ 4xy ϩ 4y2)
ϭ 3(x ϩ 2y)2
2. Apply the perfect-square pattern.
3. Factor a trinomial of the form
x2 ϩ bx ϩ c into the product of two
binomials.
4. Factor a trinomial of the form
ax2 ϩ bx ϩ c into the product of two
binomials.
Use factoring to solve
equations.
(Section 6.1/Objective 4;
Section 6.2/Objective 2;
Section 6.3/Objective 2;
Section 6.4/Objective 2;
Section 6.5/Objective 3)
Property 6.1 states that for all real numbers
a and b, ab ϭ 0 if and only if a ϭ 0 or
b ϭ 0. To solve equations by applying this
property, first set the equation equal to zero.
Proceed by factoring the other side of the
equation. Then set each factor equal to 0
and solve the equations.
Solve 2x2 ϭ Ϫ10x.
Solution
2x2 ϭ Ϫ10x
2x2 ϩ 10x ϭ 0
2x(x ϩ 5) ϭ 0
2x ϭ 0
or
xϩ5ϭ0
xϭ0
or
x ϭ Ϫ5
The solution is {Ϫ5, 0}.
(continued)
266
Chapter 6 • Factoring, Solving Equations, and Problem Solving
OBJECTIVE
SUMMARY
EXAMPLE
Solve word problems that
involve factoring.
Knowledge of factoring has expanded
the techniques available for solving word
problems. This chapter introduced the
Pythagorean theorem. The theorem pertains
to right triangles and states that in any right
triangle, the square of the longest side is
equal to the sum of the squares of the other
two sides. The formula for the theorem is
written as a2 ϩ b2 ϭ c2, where a and b are
the legs of the triangle and c is the
hypotenuse.
The length of one leg of a right triangle is
1 inch more than the length of the other
leg. The length of the hypotenuse is
5 inches. Find the length of the two legs.
(Section 6.3/Objectives 3 and
4; Section 6.5/Objective 4)
Solution
Let x represent the length of one leg of
the triangle. Then x ϩ 1 will represent the
length of the other leg. We know the
hypotenuse is equal to 5. Apply
Pythagorean theorem.
x2 ϩ (x ϩ 1)2 ϭ 5
x2 ϩ x2 ϩ 2x ϩ 1 ϭ 25
2x2 ϩ 2x Ϫ 24 ϭ 0
2(x2 ϩ x Ϫ 12) ϭ 0
x2 ϩ x Ϫ 12 ϭ 0
(x ϩ 4)(x Ϫ 3) ϭ 0
xϩ4ϭ0
or
xϪ3ϭ0
xϭϪ4
or
xϭ3
Because the length of a side of the triangle cannot be negative, the only viable
answer is 3. Therefore, the length of one
leg of the triangle is 3 inches, and the
length of the other leg is 4 inches.
Chapter 6 Review Problem Set
For Problems 1– 24, factor completely. Indicate any polynomials that are not factorable using integers.
1. x Ϫ 9x ϩ 14
2. 3x ϩ 21x
3. 9x Ϫ 4
4. 4x2 ϩ 8x Ϫ 5
5. 25x Ϫ 60x ϩ 36
6. n ϩ 13n ϩ 40n
7. y ϩ 11y Ϫ 12
8. 3xy2 ϩ 6x2y
2
2
2
2
2
9. x Ϫ 1
3
2
10. 18n ϩ 9n Ϫ 5
4
2
11. x2 ϩ 7x ϩ 24
12. 4x2 Ϫ 3x Ϫ 7
13. 3n2 ϩ 3n Ϫ 90
14. x3 Ϫ xy2
15. 2x ϩ 3xy Ϫ 2y
2
2
16. 4n Ϫ 6n Ϫ 40
2
17. 5x ϩ 5y ϩ ax ϩ ay
27. 2x2 ϩ 3x Ϫ 20 ϭ 0
28. 9n2 ϩ 21n Ϫ 8 ϭ 0
29. 6n2 ϭ 24
30. 16y2 ϩ 40y ϩ 25 ϭ 0
31. t3 Ϫ t ϭ 0
32. 28x2 ϩ 71x ϩ 18 ϭ 0
33. x2 ϩ 3x Ϫ 28 ϭ 0
34. (x Ϫ 2)(x ϩ 2) ϭ 21
35. 5n2 ϩ 27n ϭ 18
36. 4n2 ϩ 10n ϭ 14
37. 2x3 Ϫ 8x ϭ 0
18. 21t Ϫ 5t Ϫ 4
19. 2x Ϫ 2x
38. x2 Ϫ 20x ϩ 96 ϭ 0
20. 3x Ϫ 108x
21. 16x ϩ 40x ϩ 25
39. 4t2 ϩ 17t Ϫ 15 ϭ 0
2
3
3
2
22. xy Ϫ 3x Ϫ 2y ϩ 6
23. 15x2 Ϫ 7xy Ϫ 2y2
40. 3(x ϩ 2) Ϫ x(x ϩ 2) ϭ 0
24. 6n4 Ϫ 5n3 ϩ n2
41. (2x Ϫ 5)(3x ϩ 7) ϭ 0
For Problems 25– 44, solve each equation.
42. (x ϩ 4)(x Ϫ 1) ϭ 50
25. x ϩ 4x Ϫ 12 ϭ 0
43. Ϫ7n Ϫ 2n2 ϭ Ϫ15
2
26. x ϭ 11x
2