Lesson 8-5 Quadratic Equations: Differences of Squares
Differences of Squares – If two perfect squares are subtracted, then they are called the
“Difference of Squares”. (a+b) (a-b )= a2 - b2
Example 1 Factor Differences of Squares
Factor each polynomial.
a. a2 – 49
2
2
a2 – 49 = a2 – 72
Write in the form a – b .
= (a + 7)(a – 7)
Factor the difference of squares.
b. 64m2 – n2
64m2 – n2 = (8m)2 – (n)2
= (8m + n)(8m2 – n)
2
2
Write in the form a – b .
Factor the difference of squares.
c. 5x2y – 500y
Because the terms have a common factor, factor out the GCF first. Then proceed
with other factoring techniques.
5x2y – 500y = 5y(x2 – 100)
Factor out the GCF of 5y.
2
2
= 5y[(x)2 – (10)2]
Write in the form a – b .
= 5y(x + 10)(x – 10)
Factor the difference of squares.
Example 2 Apply a Technique More than Once
Factor each polynomial.
a. 16 – m4
4
2
2
16 – m4 = (4)2 – (m2)2
Write 16 – m in a – b form.
2
2
= (4 + m )(4 – m )
Factor the difference of squares.
Notice that the factor 4 – m2 is also the difference of squares.
2
2
2
= (4 + m2)(22 – m2)
Write 4 – m in a – b form.
2
= (4 + m )(2 – m)(2 + m) Factor the difference of squares.
b. 256x4 – 1
256x4 – 1 = (16x2)2 – (1)2
= (16x2 + 1)(16x2 – 1)
= (16x2 + 1)[(4x)2 – 12]
= (16x2 + 1)(4x2 + 1)(4x – 1)
4
2
2
Write 256x – 1 in a – b form.
Factor the difference of squares.
2
2
2
Write 16x – 1 in a – b form.
Factor the difference of squares.
Example 3 Apply Different Techniques
Factor each polynomial.
a. 4w4 – 144
4w4 – 144 = 4(w4 – 36)
Factor out GCF.
4
2
2
= 4[(w2)2 – (6)2]
Write w – 36 in the form a – b .
2
2
= 4(w – 6)(w + 6)
Factor the difference of squares.
Notice that the factor w2 – 6 is not the difference of squares because 6 is not a perfect square.
b. 2a3 – 10a2 – 8a + 40
2a3 – 10a2 – 8a + 40
= 2(a3 – 5a2 – 4a + 20)
= 2[(a3 – 5a2) – (4a – 20)]
= 2[a2(a – 5) – 4(a – 5)]
= 2[(a – 5)(a2 – 4)]
= 2(a – 5)(a + 2)(a – 2)
Original expression
Factor out the GCF.
Group terms with common factors.
Factor each grouping.
a – 5 is the common factor.
Factor the difference of squares.
Standardized Test Example 4 Solve Equations by Factoring
In the equation x3 + 2x2 – x – 2 = y, which is a value of x when y = 0?
1
A –1
B 0
C 2
D 3
Read the Test Item
Factor x3 + 2x2 – x – 2. Then find the values of x.
Solve the Test Item
x3 + 2x2 – x – 2 = y
x3 + 2x2 – x – 2 = 0
(x3 + 2x2) + (–x – 2) = 0
2
x (x + 2) + (–1)( x + 2) = 0
(x + 2)(x2 – 1) = 0
(x + 2)(x + 1)(x – 1) = 0
Original equation
Replace y with 0.
Group terms with a common factor.
Factor each grouping.
x + 2 is the common factor.
Factor the difference of squares.
x + 2 = 0 or x + 1 = 0 or x – 1 = 0
x = –2
x = –1
x=1
Solve each equation.
The solutions are –2, –1, and 1. The correct answer is A.