Copyright © Cengage Learning. All rights reserved. Factoring Polynomials and Solving Equations by Factoring 5

Copyright © Cengage Learning. All rights reserved. Section 5.2 Factoring the Difference of Two Squares

3 Objectives Factor the difference of two squares. Completely factor a polynomial

4 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 x 2 – y 2. (x + y)(x – y) = x 2 – xy + xy – y 2 = x 2 – y 2 In this section, we will show how to reverse the multiplication process and factor binomials such as x 2 – y 2 into binomial conjugates.

5 Factor the difference of two squares 1.

6 Factor the difference of two squares The binomial x 2 – y 2 is called the difference of two squares, because x 2 is the square of x and y 2 is the square of y. The difference of the squares of two quantities always factors into binomial conjugates. Factoring the Difference of Two Squares x 2 – y 2 = (x + y)(x – y)

7 Factor the difference of two squares To factor x 2 – 9, we note that it can be written in the form x 2 – 3 2. x 2 – 3 2 = (x + 3)(x – 3) We can check by verifying that (x + 3)(x – 3) = x 2 – 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 20 2 = 400.

8 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 x 4 y 2 are also perfect squares, because they can be written as the square of a quantity: x 4 y 2 = (x 2 y) 2

9 Example Factor: 25x 2 – 49 Solution: We can write 25x 2 – 49 in the form (5x) 2 – x 2 – 49 = (5x) 2 – 7 2 = (5x + 7)(5x – 7) We can check by multiplying (5x + 7) and (5x – 7). (5x + 7)(5x – 7) = 25x 2 – 35x + 35x – 49 = 25x 2 – 49

10 Completely factor a polynomial 2.

11 Completely factor a polynomial We often must factor out a greatest common factor before factoring the difference of two squares. To factor 8x 2 – 32, for example, we factor out the GCF of 8 and then factor the resulting difference of two squares. 8x 2 – 32 = 8(x 2 – 4) = 8(x 2 – 2 2 ) = 8(x + 2)(x – 2) Factor out 8, the GCF. Write 4 as 2 2. Factor the difference of two squares.

12 Completely factor a polynomial We can check by multiplication: 8(x + 2)(x – 2) = 8(x 2 – 4) = 8x 2 – 32

13 Example Factor completely: 2a 2 x 3 y – 8b 2 xy. Solution: We factor out the GCF of 2xy and then factor the resulting difference of two squares. 2a 2 x 3 y – 8b 2 xy = 2xy  a 2 x 2 – 2xy  4b 2 = 2xy(a 2 x 2 – 4b 2 ) The GCF is 2xy. Factor out 2xy.

14 Example – Solution = 2xy[(ax) 2 – (2b) 2 ] = 2xy(ax + 2b)(ax – 2b) Check by multiplication. Write a 2 x 2 as (ax) 2 and 4b 2 as (2b) 2. Factor the difference of two squares. cont’d

15 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 625a 4 – 81b 4 can be written in the form (25a 2 ) 2 – (9b 2 ) 2, which factor as 625a 4 – 81b 4 = (25a 2 ) 2 – (9b 2 ) 2 = (25a 2 + 9b 2 )(25a 2 – 9b 2 )

16 Completely factor a polynomial Since the factor 25a 2 – 9b 2 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, 625a 4 – 81b 4 = (25a 2 + 9b 2 )(5a + 3b)(5a – 3b)

17 Completely factor a polynomial The binomial 25a 2 + 9b 2 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.

18 Your Turn Factor completely 1.m 4 – 16n 4 o m 4 – 2 4 n 4 m 4 – (2n) 4 (m 2 + (2n) 2 )(m 2 – (2n) 2 ) (m 2 + 4n 2 )(m + 2n)(m – 2n) 2.a 5 – ab 4 o a(a 4 – b 4 ) a(a 2 + b 2 )(a 2 – b 2 ) a(a 2 + b 2 )(a + b)(a – b)

19 Your Turn Factor completely. 3.2x 8 y 2 – 32y 6 3.2y 2 (x 8 – 16y 4 ) 2y 2 ((x 4 ) 2 – (4y 2 ) 2 ) 2y 2 (x 4 + 4y 2 )(x 4 – 4y 2 ) 2y 2 (x 4 + 4y 2 )(x 2 + 2y)(x 2 – 2y) 4.x 8 y 8 – 1 o (x 4 y 4 ) 2 – 1 (x 4 y 4 + 1)(x 4 y 4 – 1) (x 4 y 4 + 1)(x 2 y 2 + 1)(x 2 y 2 – 1) (x 4 y 4 + 1)(x 2 y 2 + 1)(xy + 1)(xy – 1)

20 Your Turn 5.2a 3 b – 242ab o 2ab(a 2 – 121) 2ab(a 2 – 11 2 ) 2ab(a + 11)(a – 11) 6.81r 4 – 256s 4 o 9 2 r 4 – 16 2 s 4 ( 9r 2 ) 2 – (16s 2 ) 2 (9r s 2 )(9r 2 – 16s 2 ) (9r s 2 )(9r 2 – 16s 2 ) (9r s 2 )(3r + 4s)(3r – 4s)