Factoring Polynomials and
Solving Equations
by Factoring
Copyright © Cengage Learning. All rights reserved.
5
Section
5.2
Factoring the Difference of
Two Squares
Copyright © Cengage Learning. All rights reserved.
Objectives
1
Factor the difference of two squares.
2
Completely factor a polynomial.
3
Factoring the Difference of Two Squares
Whenever we multiply binomial conjugates, binomials of
the form (x + y) and (x – y), we obtain a binomial of the
form x2 – y2.
(x + y)(x – y) = x2 – xy + xy – y2
= x2 – y2
In this section, we will show how to reverse the
multiplication process and factor binomials such as x2 – y2
into binomial conjugates.
4
1.
Factor the difference of two
squares
5
Factor the difference of two squares
The binomial x2 – y2 is called the difference of two
squares, because x2 is the square of x and y2 is the square
of y.
The difference of the squares of two quantities always
factors into binomial conjugates.
Factoring the Difference of Two Squares
x2 – y2 = (x + y)(x – y)
6
Factor the difference of two squares
To factor x2 – 9, we note that it can be written in the form
x2 – 32.
x2 – 32 = (x + 3)(x – 3)
We can check by verifying that (x + 3)(x – 3) = x2 – 9.
To factor the difference of two squares, it is helpful to know
the integers that are perfect squares. The number 400, for
example, is a perfect square, because 202 = 400.
7
Factor the difference of two squares
The integer squares less than 400 are
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225,
256, 289, 324, 361
Expressions containing variables such as x4y2 are also
perfect squares, because they can be written as the square
of a quantity:
x4y2 = (x2y)2
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Example
Factor: 25x2 – 49
Solution:
We can write 25x2 – 49 in the form (5x)2 – 72.
25x2 – 49 = (5x)2 – 72
= (5x + 7)(5x – 7)
We can check by multiplying (5x + 7) and (5x – 7).
(5x + 7)(5x – 7) = 25x2 – 35x + 35x – 49
= 25x2 – 49
9
2.
Completely factor a polynomial
10
Completely factor a polynomial
We often must factor out a greatest common factor before
factoring the difference of two squares.
To factor 8x2 – 32, for example, we factor out the GCF of 8
and then factor the resulting difference of two squares.
8x2 – 32 = 8(x2 – 4)
Factor out 8, the GCF.
= 8(x2 – 22)
Write 4 as 22.
= 8(x + 2)(x – 2)
Factor the difference of two squares.
11
Completely factor a polynomial
We can check by multiplication:
8(x + 2)(x – 2) = 8(x2 – 4)
= 8x2 – 32
12
Example
Factor completely: 2a2x3y – 8b2xy.
Solution:
We factor out the GCF of 2xy and then factor the resulting
difference of two squares.
2a2x3y – 8b2xy
= 2xy  a2x2 – 2xy  4b2
The GCF is 2xy.
= 2xy(a2x2 – 4b2)
Factor out 2xy.
13
Example – Solution
cont’d
= 2xy[(ax)2 – (2b)2]
Write a2x2 as (ax)2 and 4b2 as (2b)2.
= 2xy(ax + 2b)(ax – 2b)
Factor the difference of two squares.
Check by multiplication.
14
Completely factor a polynomial
Sometimes we must factor a difference of two squares
more than once to completely factor a polynomial.
For example, the binomial 625a4 – 81b4 can be written in
the form (25a2)2 – (9b2)2, which factor as
625a4 – 81b4 = (25a2)2 – (9b2)2
= (25a2 + 9b2)(25a2 – 9b2)
15
Completely factor a polynomial
Since the factor 25a2 – 9b2 can be written in the form
(5a)2 – (3b)2, it is the difference of two squares and can be
factored as (5a + 3b)(5a – 3b).
Thus,
625a4 – 81b4 = (25a2 + 9b2)(5a + 3b)(5a – 3b)
16
Completely factor a polynomial
The binomial 25a2 + 9b2 is the sum of two squares,
because it can be written in the form (5a)2 + (3b)2.
If we are limited to rational coefficients, binomials that are
the sum of two squares cannot be factored unless they
contain a GCF.
Polynomials that do not factor are called prime
polynomials.
17
Your Turn
Factor completely
1. m4 – 16n4
o m4 – 24n4
m4 – (2n)4
(m2 + (2n)2)(m2 – (2n)2)
(m2 + 4n2)(m + 2n)(m – 2n)
2. a5 – ab4
o a(a4 – b4)
a(a2 + b2)(a2 – b2)
a(a2 + b2)(a + b)(a – b)
18
Your Turn
Factor completely.
3.2x8y2 – 32y6
3. 2y2(x8 – 16y4)
2y2((x4)2 – (4y2)2)
2y2(x4 + 4y2)(x4 – 4y2)
2y2(x4 + 4y2)(x2 + 2y)(x2 – 2y)
4.x8y8 – 1
o (x4y4)2 – 1
(x4y4 + 1)(x4y4 – 1)
(x4y4 + 1)(x2y2 + 1)(x2y2 – 1)
(x4y4 + 1)(x2y2 + 1)(xy + 1)(xy – 1)
19
Your Turn
5. 2a3b – 242ab
o 2ab(a2 – 121)
2ab(a2 – 112)
2ab(a + 11)(a – 11)
6. 81r4 – 256s4
o 92r4 – 162s4
(9r2)2 – (16s2)2
(9r2 + 16s2)(9r2 – 16s2)
(9r2 + 16s2)(9r2 – 16s2)
(9r2 + 16s2)(3r + 4s)(3r – 4s)
20