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2: Factoring the Difference of Two Squares

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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.

238

Chapter 6 • Factoring, Solving Equations, and Problem Solving

Classroom Example

Factor completely y 4 Ϫ 81.

EXAMPLE 3

Factor completely x4 Ϫ 16.

Solution

x4 Ϫ 16 ϭ (x2 ϩ 4)(x2 Ϫ 4)

ϭ (x2 ϩ 4)(x ϩ 2)(x Ϫ 2)

The following examples should help you to summarize the factoring ideas presented

thus far.

5x2 ϩ 20 ϭ 5(x2 ϩ 4)

25 Ϫ y2 ϭ (5 Ϫ y)(5 ϩ y)

3 Ϫ 3x2 ϭ 3(1 Ϫ x2 ) ϭ 3(1 ϩ x)(1 Ϫ x)

36x2 Ϫ 49y2 ϭ (6x Ϫ 7y)(6x ϩ 7y)

a2 ϩ 9 is not factorable using integers

9x ϩ 17y is not factorable using integers

Solving Equations

Each time we learn a new factoring technique, we also develop more tools for solving equations. Let’s consider how we can use the difference-of-squares factoring pattern to help solve

certain kinds of equations.

Classroom Example

Solve x2 ϭ 49.

EXAMPLE 4

Solve x2 ϭ 25.

Solution

x2 ϭ 25

x2 Ϫ 25 ϭ 0

(x ϩ 5)(x Ϫ 5) ϭ 0

x ϩ 5 ϭ 0     or    x Ϫ 5 ϭ 0

xϭ5

x ϭ Ϫ5    or

Remember: ab ϭ 0 if and only if a ϭ 0 or b ϭ 0

The solution set is {Ϫ5, 5}. Check these answers!

Classroom Example

Solve 64a2 ϭ 121.

EXAMPLE 5

Solve 9x2 ϭ 25.

Solution

9x2 ϭ 25

9x Ϫ 25 ϭ 0

(3x ϩ 5)(3x Ϫ 5) ϭ 0

3x ϩ 5 ϭ 0

3x Ϫ 5 ϭ 0

or

3x ϭ Ϫ5

3x ϭ 5

or

5

5

xϭϪ

or

3

3

2

5 5

The solution set is eϪ , f .

3 3

6.2 • Factoring the Difference of Two Squares

Classroom Example

Solve 3x 2 ϭ 27.

EXAMPLE 6

239

Solve 5y2 ϭ 20.

Solution

5y2 ϭ 20

5y2

20

Divide both sides by 5

ϭ

5

5

y2 ϭ 4

2

y Ϫ4ϭ0

(y ϩ 2)(y Ϫ 2) ϭ 0

y ϩ 2 ϭ 0

or  y Ϫ 2 ϭ 0

y ϭ Ϫ2  or   y ϭ 2

The solution set is {Ϫ2, 2}. Check it!

Classroom Example

Solve c3 Ϫ 36c ϭ 0.

EXAMPLE 7

Solve x3 Ϫ 9x ϭ 0.

Solution

x3 Ϫ 9x ϭ 0

x(x2 Ϫ 9) ϭ 0

x(x Ϫ 3)(x ϩ 3) ϭ 0

x ϭ 0  or  x Ϫ 3 ϭ 0  or  x ϩ 3 ϭ 0

x ϭ 0  or  x ϭ 3

or  x ϭ Ϫ3

The solution set is {Ϫ3, 0, 3}.

The more we know about solving equations, the more easily we can solve word problems.

Classroom Example

The combined area of two squares is

360 square inches. Each side of one

square is three times as long as a

side of the other square. Find the

lengths of the sides of each square.

EXAMPLE 8

The combined area of two squares is 20 square centimeters. Each side of one square is twice

as long as a side of the other square. Find the lengths of the sides of each square.

Solution

We can sketch two squares and label the sides of the smaller square s (see Figure 6.2). Then

the sides of the larger square are 2s. Since the sum of the areas of the two squares is 20 square

centimeters, we can set up and solve the following equation:

s2 ϩ (2s) 2 ϭ 20

s2 ϩ 4s2 ϭ 20

5s2 ϭ 20

s2 ϭ 4

s2 Ϫ 4 ϭ 0

(s ϩ 2)(s Ϫ 2) ϭ 0

s ϩ 2 ϭ 0   or  s Ϫ 2 ϭ 0

s ϭ Ϫ2  or  s ϭ 2

s

2s

s

2s

Figure 6.2

Since s represents the length of a side of a square, we must disregard the solution Ϫ2. Thus

one square has sides of length 2 centimeters and the other square has sides of length 2(2) ϭ 4

centimeters.

240

Chapter 6 • Factoring, Solving Equations, and Problem Solving

Concept Quiz 6.2

For Problems 1–10, answer true or false.

1. A binomial that has two perfect square terms that are subtracted is called the difference

of two squares.

2. The sum of two squares is factorable using integers.

3. When factoring it is usually best to look for a common factor first.

4. The polynomial 4x2 ϩ y2 factors into (2x ϩ y)(2x ϩ y).

5. The completely factored form of y4 Ϫ 81 is (y2 ϩ 9)(y2 Ϫ 9).

6. The solution set for x2 ϭ Ϫ16 is {Ϫ4}.

7. The solution set for 5x3 Ϫ 5x ϭ 0 is {Ϫ1, 0, 1}.

8. The solution set for x4 Ϫ 9x2 ϭ 0 is {Ϫ3, 0, 3}.

9. The completely factored form of x4 Ϫ 1 is (x ϩ 1)(x Ϫ 1)(x2 ϩ 1).

10. The completely factored form of 2x3 y Ϫ 8xy is 2xy(x ϩ 2)(x Ϫ 2).

Problem Set 6.2

For Problems 1–12, use the difference-of-squares pattern to

factor each polynomial. (Objective 1)

1. x Ϫ 1

2. x Ϫ 25

3. x2 Ϫ 100

4. x2 Ϫ 121

5. x2 Ϫ 4y2

6. x2 Ϫ 36y2

7. 9x2 Ϫ y2

8. 49y2 Ϫ 64x2

2

2

41. 16x4 Ϫ 81y4

42. x4 Ϫ 1

43. 81 Ϫ x4

44. 81x4 Ϫ 16y4

For Problems 45–68, solve each equation. (Objective 2)

45. x2 ϭ 9

46. x2 ϭ 1

47. 4 ϭ n2

48. 144 ϭ n2

49. 9x2 ϭ 16

50. 4x2 ϭ 9

51. n2 Ϫ 121 ϭ 0

52. n2 Ϫ 81 ϭ 0

53. 25x2 ϭ 4

54. 49x2 ϭ 36

For Problems 13– 44, factor each polynomial completely. Indicate any that are not factorable using integers. Don’t forget

to look for a common monomial factor first. (Objective 1)

55. 3x2 ϭ 75

56. 7x2 ϭ 28

57. 3x3 Ϫ 48x ϭ 0

58. x3 Ϫ x ϭ 0

13. 5x2 Ϫ 20

14. 7x2 Ϫ 7

59. n3 ϭ 16n

60. 2n3 ϭ 8n

15. 8x2 ϩ 32

16. 12x2 ϩ 60

61. 5 Ϫ 45x2 ϭ 0

62. 3 Ϫ 12x2 ϭ 0

17. 2x2 Ϫ 18y2

18. 8x2 Ϫ 32y2

63. 4x3 Ϫ 400x ϭ 0

64. 2x3 Ϫ 98x ϭ 0

19. x3 Ϫ 25x

20. 2x3 Ϫ 2x

65. 64x2 ϭ 81

66. 81x2 ϭ 25

21. x2 ϩ 9y2

22. 18x Ϫ 42y

67. 36x3 ϭ 9x

68. 64x3 ϭ 4x

23. 45x2 Ϫ 36xy

24. 16x2 ϩ 25y2

25. 36 Ϫ 4x2

26. 75 Ϫ 3x2

27. 4a4 ϩ 16a2

28. 9a4 ϩ 81a2

29. x Ϫ 81

30. 16 Ϫ x

31. x ϩ x

32. x5 ϩ 2x3

33. 3x3 ϩ 48x

35. 5x Ϫ 20x3

34. 6x3 ϩ 24x

36. 4x Ϫ 36x3

37. 4x2 Ϫ 64

38. 9x2 Ϫ 9

39. 75x3y Ϫ 12xy3

40. 32x3y Ϫ 18xy3

9. 36a2 Ϫ 25b2

11. 1 Ϫ 4n2

4

4

2

10. 4a2 Ϫ 81b2

12. 4 Ϫ 9n2

4

For Problems 69–80, set up an equation and solve the

problem. (Objective 2)

69. Forty-nine less than the square of a number equals zero.

Find the number.

70. The cube of a number equals nine times the number.

Find the number.

71. Suppose that five times the cube of a number equals

80 times the number. Find the number.

72. Ten times the square of a number equals 40. Find the

number.

6.2 • Factoring the Difference of Two Squares

73. The sum of the areas of two squares is 234 square inches. Each side of the larger square is five times the length

of a side of the smaller square. Find the length of a side

of each square.

74. The difference of the areas of two squares is 75 square

feet. Each side of the larger square is twice the length of

a side of the smaller square. Find the length of a side of

each square.

1

75. Suppose that the length of a certain rectangle is 2 times

2

its width, and the area of that same rectangle is 160 square

centimeters. Find the length and width of the rectangle.

76. Suppose that the width of a certain rectangle is threefourths of its length, and the area of that same rectangle

is 108 square meters. Find the length and width of the

rectangle.

241

77. The sum of the areas of two circles is 80␲ square

meters. Find the length of a radius of each circle if one

of them is twice as long as the other.

78. The area of a triangle is 98 square feet. If one side of the

triangle and the altitude to that side are of equal length,

find the length.

79. The total surface area of a right circular cylinder is

100␲ square centimeters. If a radius of the base and the

altitude of the cylinder are the same length, find the

80. The total surface area of a right circular cone is square

feet. If the slant height of the cone is equal in length to a

diameter of the base, find the length of a radius.

Thoughts Into Words

81. How do we know that the equation x2 ϩ 1 ϭ 0 has no

solutions in the set of real numbers?

82. Why is the following factoring process incomplete?

16x2 Ϫ 64 ϭ (4x ϩ 8)(4x Ϫ 8)

How could the factoring be done?

83. Consider the following solution:

4x2 Ϫ 36 ϭ 0

4(x2 Ϫ 9) ϭ 0

4(x ϩ 3) (x Ϫ 3) ϭ 0

4 ϭ 0   or    x ϩ 3 ϭ 0    or    x Ϫ 3 ϭ 0

x ϭ Ϫ3   or

xϭ3

4 ϭ 0   or

The solution set is {Ϫ3, 3}. Is this a correct solution? Do

you have any suggestions to offer the person who did

this problem?

Further Investigations

The following patterns can be used to factor the sum and difference of two cubes.

a ϩ b ϭ (a ϩ b) (a Ϫ ab ϩ b )

a3 Ϫ b3 ϭ (a Ϫ b) (a2 ϩ ab ϩ b2 )

3

3

2

2

88. 8x3 ϩ 27y3

89. 27a3 Ϫ 64b3

90. 1 Ϫ 8x3

91. 1 ϩ 27a3

Consider these examples.

x3 ϩ 8 ϭ (x) 3 ϩ (2) 3 ϭ (x ϩ 2)(x2 Ϫ 2x ϩ 4)

x3 Ϫ 1 ϭ (x) 3 Ϫ (1) 3 ϭ (x Ϫ 1)(x2 ϩ x ϩ 1)

92. x3 ϩ 8y3

93. 8x3 Ϫ y3

94. a3b3 Ϫ 1

Use the sum and difference-of-cubes patterns to factor each

polynomial.

96. 8 ϩ n3

84. x3 ϩ 1

97. 125x3 ϩ 8y3

85. x3 Ϫ 8

98. 27n3 Ϫ 125

86. n3 Ϫ 27

99. 64 ϩ x3

87. n3 ϩ 64

1. True

2. False

3. True

4. False

95. 27x3 Ϫ 8y3

5. False

6. False

7. True

8. True

9. True

10. True

242

Chapter 6 • Factoring, Solving Equations, and Problem Solving

6.3

Factoring Trinomials of the Form x 2 ؉ bx ؉ c

OBJECTIVES

1

Factor trinomials of the form x2 ϩ bx ϩ c

2

Use factoring of trinomials to solve equations

3

Solve word problems involving consecutive numbers

4

Use the Pythagorean theorem to solve problems

One of the most common types of factoring used in algebra is the expression of a trinomial

as the product of two binomials. In this section we will consider trinomials for which the

coefficient of the squared term is 1; that is, trinomials of the form x2 ϩ bx ϩ c.

Again, to develop a factoring technique, we first look at some multiplication ideas.

Consider the product (x ϩ r)(x ϩ s), and use the distributive property to show how each term

of the resulting trinomial is formed.

(x ϩ r)(x ϩ s) ϭ x(x) ϩ x(s) ϩ r(x) ϩ r(s)

14243

x

2

ϩ

(s ϩ r)x

ϩ

rs

Notice that the coefficient of the middle term is the sum of r and s, and the last term is the

product of r and s. These two relationships are used in the next examples.

Classroom Example

Factor x 2 ϩ 12x ϩ 27.

EXAMPLE 1

Factor x2 ϩ 7x ϩ 12.

Solution

We need to fill in the blanks with two numbers whose product is 12 and whose sum is 7.

x2 ϩ 7x ϩ 12 ϭ (x ϩ _____)(x ϩ _____)

To assist in finding the numbers, we can set up a table of the factors of 12.

Product

Sum

1(12) ϭ 12

2(6) ϭ 12

3(4) ϭ 12

1 ϩ 12 ϭ 13

2ϩ6ϭ8

3ϩ4ϭ7

The bottom line contains the numbers that we need. Thus

x 2 ϩ 7x ϩ 12 ϭ (x ϩ 3)(x ϩ 4)

Classroom Example

Factor b2 Ϫ 11b ϩ 28.

EXAMPLE 2

Factor x 2 Ϫ 11x ϩ 24.

Solution

To factor x2 Ϫ 11x ϩ 24, we want to find two numbers whose product is 24 and whose sum

is Ϫ11.

Product

(Ϫ1)(Ϫ 24)

(Ϫ2)(Ϫ12)

(Ϫ3)(Ϫ8)

(Ϫ4)(Ϫ 6)

Sum

ϭ 24  Ϫ1 ϩ (Ϫ24)

ϭ 24 Ϫ2 ϩ (Ϫ12)

ϭ 24

Ϫ 3 ϩ (Ϫ 8)

ϭ 24

Ϫ 4 ϩ (Ϫ6)

ϭ Ϫ 25

ϭ Ϫ 14

ϭ Ϫ 11

ϭ Ϫ 10

6.3 • Factoring Trinomials of the Form x 2 ϩ bx ϩ c

243

The third line contains the numbers that we want. Thus

x2 Ϫ 11x ϩ 24 ϭ (x Ϫ 3) (x Ϫ 8)

Classroom Example

Factor x2 ϩ 2x Ϫ 24.

EXAMPLE 3

Factor x2 ϩ 3x Ϫ 10.

Solution

To factor x2 ϩ 3x Ϫ10, we want to find two numbers whose product is Ϫ10 and whose sum is 3.

Product

1(Ϫ10)

Ϫ1(10)

2(Ϫ5)

Ϫ 2(5)

ϭ Ϫ10

ϭ Ϫ 10

ϭ Ϫ 10

ϭ Ϫ 10

Sum

1 ϩ (Ϫ10) ϭ Ϫ9

Ϫ1 ϩ 10 ϭ 9

2 ϩ (Ϫ5) ϭ Ϫ 3

Ϫ2 ϩ 5 ϭ 3

The bottom line is the key line. Thus

x2 ϩ 3x Ϫ 10 ϭ (x ϩ 5)(x Ϫ 2)

Classroom Example

Factor n2 Ϫ 4n Ϫ 32.

EXAMPLE 4

Factor x2 Ϫ 2x Ϫ 8.

Solution

We are looking for two numbers whose product is Ϫ8 and whose sum is Ϫ2.

Product

1(Ϫ8)

Ϫ1(8)

2(Ϫ4)

Ϫ 2(4)

ϭ Ϫ8

ϭ Ϫ8

ϭ Ϫ8

ϭ Ϫ8

Sum

1 ϩ (Ϫ8) ϭ Ϫ7

Ϫ1 ϩ 8 ϭ 7

2 ϩ (Ϫ4) ϭ Ϫ 2

Ϫ2 ϩ 4 ϭ 2

The third line has the information we want.

x2 Ϫ 2x Ϫ 8 ϭ (x Ϫ 4)(x ϩ 2)

The tables in the last four examples illustrate one way of organizing your thoughts for such

problems. We show complete tables; that is, for Example 4, we include the bottom line even

though the desired numbers are obtained in the third line. If you use such tables, keep in mind

that as soon as you get the desired numbers, the table need not be completed any further.

Furthermore, you may be able to find the numbers without using a table. The key ideas are the

product and sum relationships.

Classroom Example

Factor x2 ϩ 13x Ϫ 14.

EXAMPLE 5

Factor x2 Ϫ 13x ϩ 12.

Solution

Product

(Ϫ1)(Ϫ12) ϭ 12

Sum

(Ϫ1) ϩ (Ϫ12) ϭ Ϫ13

We need not complete the table.

x2 Ϫ 13x ϩ 12 ϭ (x Ϫ 1)(x Ϫ 12)

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