2: Factoring the Difference of Two Squares
Tải bản đầy đủ - 0trang
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
Added Ϫ25 to both sides
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
xϭ
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
length of a radius.
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
Answers to the Concept Quiz
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)