1: Factoring by Using the Distributive Property
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6.1 • Factoring by Using the Distributive Property
231
The next four examples further illustrate this process of factoring out the greatest common
monomial factor.
Classroom Example
Factor 18x 4 Ϫ 24x 2.
EXAMPLE 3
Factor 12x3 Ϫ 8x2.
Solution
12x3 Ϫ 8x2 ϭ 4x2 (3x) Ϫ 4x2 (2)
ϭ 4x2 (3x Ϫ 2)
Classroom Example
Factor 15ab3 ϩ 27a2b.
EXAMPLE 4
ab Ϫ ac ϭ a(b Ϫ c)
Factor 12x2y ϩ 18x y2.
Solution
12x2y ϩ 18xy2 ϭ 6xy(2x) ϩ 6xy(3y)
ϭ 6xy(2x ϩ 3y)
Classroom Example
Factor 12k 2 Ϫ 33k 3 ϩ 51k 5.
EXAMPLE 5
Factor 24x3 ϩ 30x4 Ϫ 42x5.
Solution
24x3 ϩ 30x4 Ϫ 42x5 ϭ 6x3 (4) ϩ 6x3 (5x) Ϫ 6x3 (7x2 )
ϭ 6x3 (4 ϩ 5x Ϫ 7x2 )
Classroom Example
Factor 5y 4 ϩ 5y 3.
EXAMPLE 6
Factor 9x2 ϩ 9x.
Solution
9x2 ϩ 9x ϭ 9x(x) ϩ 9x(1)
ϭ 9x(x ϩ 1)
We want to emphasize the point made just before Example 3. It is important to realize
that we are factoring out the greatest common monomial factor. We could factor an expression such as 9x2 ϩ 9x in Example 6 as 9(x2 ϩ x) , 3(3x2 ϩ 3x) , 3x(3x ϩ 3) , or even
1
(18x2 ϩ 18x) , but it is the form 9x(x ϩ 1) that we want. We can accomplish this by factoring
2
out the greatest common monomial factor; we sometimes refer to this process as factoring
completely. A polynomial with integral coefficients is in completely factored form if these
conditions are met:
1. It is expressed as a product of polynomials with integral coefficients.
2. No polynomial, other than a monomial, within the factored form can be further factored into polynomials with integral coefficients.
Thus 9(x2 ϩ x), 3(3x2 ϩ 3x), and 3x(3x ϩ 3) are not completely factored because they violate
1
condition 2. The form (18x2 ϩ 18x) violates both conditions 1 and 2.
2
Sometimes there may be a common binomial factor rather than a common monomial
factor. For example, each of the two terms of x(y ϩ 2) ϩ z(y ϩ 2) has a binomial factor of
(y ϩ 2). Thus we can factor (y ϩ 2) from each term and get
x(y ϩ 2) ϩ z(y ϩ 2) ϭ (y ϩ 2)(x ϩ z)
232
Chapter 6 • Factoring, Solving Equations, and Problem Solving
Consider a few more examples involving a common binomial factor.
a(b ϩ c) Ϫ d(b ϩ c) ϭ (b ϩ c) (a Ϫ d)
x(x ϩ 2) ϩ 3(x ϩ 2) ϭ (x ϩ 2) (x ϩ 3)
x(x ϩ 5) Ϫ 4(x ϩ 5) ϭ (x ϩ 5) (x Ϫ 4)
It may be that the original polynomial exhibits no apparent common monomial or binomial factor, which is the case with
ab ϩ 3a ϩ bc ϩ 3c
However, by factoring a from the first two terms and c from the last two terms, we see that
ab ϩ 3a ϩ bc ϩ 3c ϭ a(b ϩ 3) ϩ c(b ϩ 3)
Now a common binomial factor of (b ϩ 3) is obvious, and we can proceed as before.
a(b ϩ 3) ϩ c(b ϩ 3) ϭ (b ϩ 3)(a ϩ c)
This factoring process is called factoring by grouping. Let’s consider two more examples of
factoring by grouping.
x2 Ϫ x ϩ 5x Ϫ 5 ϭ x(x Ϫ 1) ϩ 5(x Ϫ 1)
ϭ (x Ϫ 1)(x ϩ 5)
Factor x from first two terms and 5 from
last two terms
Factor common binomial factor of
(x Ϫ 1) from both terms
6x2 Ϫ 4x Ϫ 3x ϩ 2 ϭ 2x(3x Ϫ 2) Ϫ 1(3x Ϫ 2) Factor 2x from first two terms and Ϫ1
from last two terms
ϭ (3x Ϫ 2) (2x Ϫ 1)
Factor common binomial factor of
(3x Ϫ 2) from both terms
Back to Solving Equations
Suppose we are told that the product of two numbers is 0. What do we know about the numbers? Do you agree with our conclusion that at least one of the numbers must be 0? The next
property formalizes this idea.
Property 6.1
For all real numbers a and b,
ab ϭ 0 if and only if a ϭ 0 or b ϭ 0
Property 6.1 provides us with another technique for solving equations.
Classroom Example
Solve m2 Ϫ 8m ϭ 0.
EXAMPLE 7
Solve x2 ϩ 6x ϭ 0.
Solution
To solve equations by applying Property 6.1, one side of the equation must be a product, and
the other side of the equation must be zero. This equation already has zero on the right-hand
side of the equation, but the left-hand side of this equation is a sum. We will factor the left-hand
side, x 2 ϩ 6x, to change the sum into a product.
x2 ϩ 6x ϭ 0
x(x ϩ 6) ϭ 0
x ϭ 0 or x ϩ 6 ϭ 0
x ϭ Ϫ6
x ϭ 0 or
Factor
ab ϭ 0 if and only if a ϭ 0 or b ϭ 0
The solution set is {Ϫ6, 0}. (Be sure to check both values in the original equation.)
6.1 • Factoring by Using the Distributive Property
Classroom Example
Solve x2 ϭ 17x.
EXAMPLE 8
233
Solve x2 ϭ 12x.
Solution
In order to solve this equation by Property 6.1, we will first get zero on the right-hand side of
the equation by adding Ϫ12x to each side. Then we factor the expression on the left-hand side
of the equation.
x2 ϭ 12x
x Ϫ 12x ϭ 0
x(x Ϫ 12) ϭ 0
x ϭ 0 or x Ϫ 12 ϭ 0
x ϭ 0 or x ϭ 12
2
Added Ϫ12x to both sides
ab ϭ 0 if and only if a ϭ 0 or b ϭ 0
The solution set is {0, 12}.
Remark: Notice in Example 8 that we did not divide both sides of the original equation by x.
This would cause us to lose the solution of 0.
Classroom Example
Solve 5d 2 Ϫ 7d ϭ 0.
EXAMPLE 9
Solve 4x2 Ϫ 3x ϭ 0.
Solution
4x2 Ϫ 3x ϭ 0
x(4x Ϫ 3) ϭ 0
x ϭ 0 or 4x Ϫ 3 ϭ 0
x ϭ 0 or
4x ϭ 3
3
x ϭ 0 or
x ϭ
4
ab ϭ 0 if and only if a ϭ 0 or b ϭ 0
3
The solution set is e0, f .
4
Classroom Example
Solve w(w Ϫ 8)ϩ11(w Ϫ 8) ϭ 0.
EXAMPLE 10
Solve x(x ϩ 2) ϩ 3(x ϩ 2) ϭ 0.
Solution
In order to solve this equation by Property 6.1, we will factor the left-hand side of the equation. The greatest common factor of the terms is (x + 2).
x(x ϩ 2) ϩ 3(x ϩ 2) ϭ 0
(x ϩ 2)(x ϩ 3) ϭ 0
x ϩ 2 ϭ 0 or x ϩ 3 ϭ 0
x ϭ Ϫ3
x ϭ Ϫ2 or
ab ϭ 0 if and only if a ϭ 0 or b ϭ 0
The solution set is {Ϫ3, Ϫ2} .
Each time we expand our equation-solving capabilities, we also acquire more techniques for solving problems. Let’s solve a geometric problem with the ideas we learned in
this section.
234
Chapter 6 • Factoring, Solving Equations, and Problem Solving
EXAMPLE 11
Classroom Example
The area of a square is numerically
equal to three times its perimeter.
Find the length of a side of the
square.
The area of a square is numerically equal to twice its perimeter. Find the length of a side of
the square.
Solution
Sketch a square and let s represent the length of each side (see Figure 6.1). Then the area is
represented by s2 and the perimeter by 4s. Thus
s
s
s2 ϭ 2(4s)
s2 ϭ 8s
s2 Ϫ 8s ϭ 0
s(s Ϫ 8) ϭ 0
s ϭ 0 or s Ϫ 8 ϭ 0
s
s
Figure 6.1
s ϭ 0 or
sϭ8
Since 0 is not a reasonable answer to the problem, the solution is 8. (Be sure to check this
solution in the original statement of the problem!)
Concept Quiz 6.1
For Problems 1–10, answer true or false.
1. The greatest common factor of 6x2y3 Ϫ 12x3y2 ϩ 18x4y is 2x2y.
2. If the factored form of a polynomial can be factored further, then it has not met the conditions to be considered “factored completely.”
3. Common factors are always monomials.
4. If the product of x and y is zero, then x is zero or y is zero.
5. The factored form 3a(2a2 ϩ 4) is factored completely.
6. The solutions for the equation x(x ϩ 2) ϭ 7 are 7 and 5.
7. The solution set for x2 ϭ 7x is {7}.
8. The solution set for x(x Ϫ 2) Ϫ 3(x Ϫ 2) ϭ 0 is {2, 3}.
9. The solution set for Ϫ3x ϭ x2 is {Ϫ3, 0}.
10. The solution set for x(x ϩ 6) ϭ 2(x ϩ 6) is {Ϫ6}.
Problem Set 6.1
For Problems 1–10, find the greatest common factor of the
given expressions. (Objective 1)
For Problems 11– 46, factor each polynomial completely.
(Objective 2)
11. 8x ϩ 12y
12. 18x ϩ 24y
13. 14xy Ϫ 21y
14. 24x Ϫ 40xy
15. 18x ϩ 45x
16. 12x ϩ 28x3
7. 6x3, 8x, and 24x2
17. 12xy2 Ϫ 30x2y
18. 28x2y2 Ϫ 49x2y
8. 72xy, 36x2y, and 84xy2
19. 36a2b Ϫ 60a3b4
20. 65ab3 Ϫ 45a2b2
9. 16a2b2, 40a2b3, and 56a3b4
21. 16xy3 ϩ 25x2y2
22. 12x2y2 ϩ 29x2y
23. 64ab Ϫ 72cd
24. 45xy Ϫ 72zw
1. 24y and 30xy
2. 32x and 40xy
3. 60x2y and 84xy2
4. 72x3 and 63x2
3
2 2
5. 42ab and 70a b
10. 70a3b3, 42a2b4, and 49ab5
2 2
6. 48a b and 96ab
4
2
6.1 • Factoring by Using the Distributive Property
25. 9a2b4 Ϫ 27a2b
26. 7a3b5 Ϫ 42a2b6
63. x2 ϩ x ϭ 0
64. x2 ϩ 7x ϭ 0
27. 52x4y2 ϩ 60x6y
28. 70x5y3 Ϫ 42x8y2
65. n2 ϭ 5n
66. n2 ϭ Ϫ2n
29. 40x2y2 ϩ 8x2y
30. 84x2y3 ϩ 12xy3
67. 2y2 Ϫ 3y ϭ 0
68. 4y2 Ϫ 7y ϭ 0
31. 12x ϩ 15xy ϩ 21x2
69. 7x2 ϭ Ϫ3x
70. 5x2 ϭ Ϫ2x
32. 30x2y ϩ 40xy ϩ 55y
71. 3n2 ϩ 15n ϭ 0
72. 6n2 Ϫ 24n ϭ 0
33. 2x3 Ϫ 3x2 ϩ 4x
73. 4x2 ϭ 6x
74. 12x2 ϭ 8x
34. x4 ϩ x3 ϩ x2
75. 7x Ϫ x2 ϭ 0
76. 9x Ϫ x2 ϭ 0
77. 13x ϭ x2
78. 15x ϭ Ϫx2
79. 5x ϭ Ϫ2x2
80. 7x ϭ Ϫ5x2
35. 44y5 Ϫ 24y3 Ϫ 20y2
36. 14a Ϫ 18a Ϫ 26a
3
5
37. 14a b ϩ 35ab Ϫ 49a b
2 3
2
3
38. 24a3b2 ϩ 36a2b4 Ϫ 60a4b3
39. x(y ϩ 1) ϩ z(y ϩ 1)
235
81. x(x ϩ 5) Ϫ 4(x ϩ 5) ϭ 0
82. x(3x Ϫ 2) Ϫ 7(3x Ϫ 2) ϭ 0
83. 4(x Ϫ 6) Ϫ x(x Ϫ 6) ϭ 0
40. a(c ϩ d) ϩ 2(c ϩ d)
84. x(x ϩ 9) ϭ 2(x ϩ 9)
41. a(b Ϫ 4) Ϫ c(b Ϫ 4)
42. x(y Ϫ 6) Ϫ 3(y Ϫ 6)
For Problems 85–91, set up an equation and solve each problem. (Objective 4)
43. x(x ϩ 3) ϩ 6(x ϩ 3)
44. x(x Ϫ 7) ϩ 9(x Ϫ 7)
85. The square of a number equals nine times that number.
Find the number.
45. 2x(x ϩ 1) Ϫ 3(x ϩ 1)
46. 4x(x ϩ 8) Ϫ 5(x ϩ 8)
86. Suppose that four times the square of a number equals
20 times that number. What is the number?
For Problems 47–60, use the process of factoring by grouping to factor each polynomial. (Objective 3)
47. 5x ϩ 5y ϩ bx ϩ by
88. The area of a square is 14 times as large as the area of a
triangle. One side of the triangle is 7 inches long, and the
altitude to that side is the same length as a side of the
square. Find the length of a side of the square. Also find
the areas of both figures, and be sure that your answer
checks.
48. 7x ϩ 7y ϩ zx ϩ zy
49. bx Ϫ by Ϫ cx ϩ cy
50. 2x Ϫ 2y Ϫ ax ϩ ay
51. ac ϩ bc ϩ a ϩ b
52. x ϩ y ϩ ax ϩ ay
89. Suppose that the area of a circle is numerically equal to
the perimeter of a square, and that the length of a radius
of the circle is equal to the length of a side of the square.
Find the length of a side of the square. Express your answer in terms of .
53. x ϩ 5x ϩ 12x ϩ 60
2
54. x2 ϩ 3x ϩ 7x ϩ 21
55. x2 Ϫ 2x Ϫ 8x ϩ 16
56. x2 Ϫ 4x Ϫ 9x ϩ 36
90. One side of a parallelogram, an altitude to that side, and
one side of a rectangle all have the same measure. If an
adjacent side of the rectangle is 20 centimeters long, and
the area of the rectangle is twice the area of the parallelogram, find the areas of both figures.
57. 2x2 ϩ x Ϫ 10x Ϫ 5
58. 3x2 ϩ 2x Ϫ 18x Ϫ 12
59. 6n2 Ϫ 3n Ϫ 8n ϩ 4
60. 20n2 ϩ 8n Ϫ 15n Ϫ 6
For Problems 61–84, solve each equation. (Objective 4)
61. x2 Ϫ 8x ϭ 0
87. The area of a square is numerically equal to five times its
perimeter. Find the length of a side of the square.
62. x2 Ϫ 12x ϭ 0
91. The area of a rectangle is twice the area of a square. If
the rectangle is 6 inches long, and the width of the rectangle is the same as the length of a side of the square,
find the dimensions of both the rectangle and the
square.
236
Chapter 6 • Factoring, Solving Equations, and Problem Solving
Thoughts Into Words
92. Suppose that your friend factors 24x2y ϩ 36xy like this:
24x2y ϩ 36xy ϭ 4xy(6x ϩ 9)
ϭ (4xy)(3)(2x ϩ 3)
ϭ 12xy(2x ϩ 3)
Is this correct? Would you make any suggestions for
changing her method?
93. The following solution is given for the equation
x(x Ϫ 10) ϭ 0.
x(x Ϫ 10) ϭ 0
x2 Ϫ 10x ϭ 0
x(x Ϫ 10) ϭ 0
x ϭ 0 or x Ϫ 10 ϭ 0
x ϭ 0 or
x ϭ 10
The solution set is {0,10}. Is this a correct solution?
Would you suggest any changes to the method?
Further Investigations
94. The total surface area of a right circular cylinder is
given by the formula A ϭ 2 r2 ϩ 2 rh, where r represents the radius of a base, and h represents the height of
the cylinder. For computational purposes, it may be
more convenient to change the form of the right side of
the formula by factoring it.
A ϭ 2 r2 ϩ 2 rh ϭ 2 r(r ϩ h)
Use A ϭ 2 r(r ϩ h) to find the total surface area of
22
each of the following cylinders. Use
as an approx7
imation for .
A ϭ P ϩ Prt ϭ P(1 ϩ rt)
Use A ϭ P(1 ϩ rt) to find the total amount of money
accumulated for each of the following investments.
(a) $100 at 8% for 2 years
(b) $200 at 9% for 3 years
(c) $500 at 10% for 5 years
(d) $1000 at 10% for 10 years
(a) r ϭ 7 centimeters and h ϭ 12 centimeters
For Problems 96–99, solve for the indicated variable.
(b) r ϭ 14 meters and h ϭ 20 meters
96. ax ϩ bx ϭ c for x
(c) r ϭ 3 feet and h ϭ 4 feet
97. b2x2 Ϫ cx ϭ 0 for x
(d) r ϭ 5 yards and h ϭ 9 yards
95. The formula A ϭ P ϩ Prt yields the total amount of
money accumulated (A) when P dollars is invested at
Answers to the Concept Quiz
1. False
2. True
3. False
4. True
9. True
10. False
6.2
r percent simple interest for t years. For computational
purposes, it may be convenient to change the right side
of the formula by factoring.
5. False
98. 5ay2 ϭ by for y
99. y ϩ ay Ϫ by Ϫ c ϭ 0 for y
6. False
7. False
8. True
Factoring the Difference of Two Squares
OBJECTIVES
1
Factor the difference of two squares
2
Solve equations by factoring the difference of two squares
In Section 5.3 we noted some special multiplication patterns. One of these patterns was
(a Ϫ b)(a ϩ b) ϭ a2 Ϫ b2
We can view this same pattern as follows:
6.2 • Factoring the Difference of Two Squares
237
Difference of Two Squares
a2 Ϫ b2 ϭ (a Ϫ b)(a ϩ b)
To apply the pattern is a fairly simple process, as these next examples illustrate. The steps
inside the box are often performed mentally.
x2 Ϫ 36 ϭ
4x2 Ϫ 25 ϭ
9x2 Ϫ 16y2 ϭ
64 Ϫ y2 ϭ
(x) 2 Ϫ (6) 2
(2x) 2 Ϫ (5) 2
(3x) 2 Ϫ (4y) 2
(8) 2 Ϫ (y) 2
ϭ
ϭ
ϭ
ϭ
(x Ϫ 6)(x ϩ 6)
(2x Ϫ 5)(2x ϩ 5)
(3x Ϫ 4y)(3x ϩ 4y)
(8 Ϫ y)(8 ϩ y)
Since multiplication is commutative, the order of writing the factors is not important. For
example, (x Ϫ 6)(x ϩ 6) can also be written as (x ϩ 6)(x Ϫ 6).
You must be careful not to assume an analogous factoring pattern for the sum of two
squares; it does not exist. For example, x2 ϩ 4 ϶ (x ϩ 2)(x ϩ 2) because (x ϩ 2) (x ϩ 2) ϭ
x2 ϩ 4x ϩ 4. We say that the sum of two squares is not factorable using integers. The
1
phrase “using integers” is necessary because x2 ϩ 4 could be written as (2x2 ϩ 8) , but such
2
factoring is of no help. Furthermore, we do not consider (1)(x2 ϩ 4) as factoring x2 ϩ 4.
It is possible that both the technique of factoring out a common monomial factor and the
pattern difference of two squares can be applied to the same polynomial. In general, it is best
to look for a common monomial factor first.
Classroom Example
Factor 7x 2 Ϫ 28.
EXAMPLE 1
Factor 2x2 Ϫ 50.
Solution
2x2 Ϫ 50 ϭ 2(x2 Ϫ 25)
ϭ 2(x Ϫ 5)(x ϩ 5)
Common factor of 2
Difference of squares
In Example 1, by expressing 2x2 Ϫ 50 as 2(x Ϫ 5)(x ϩ 5), we say that the algebraic expression has been factored completely. That means that the factors 2, x Ϫ 5, and x ϩ 5 cannot
be factored any further using integers.
Classroom Example
Factor completely 32m 3 Ϫ 50m.
EXAMPLE 2
Factor completely 18y3 Ϫ 8y.
Solution
18y3 Ϫ 8y ϭ 2y(9y2 Ϫ 4)
ϭ 2y(3y Ϫ 2)(3y ϩ 2)
Common factor of 2y
Difference of squares
Sometimes it is possible to apply the difference-of-squares pattern more than once.
Consider the next example.