Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.4 Factoring Perfect-Square Trinomials and Differences of Squares - 5.4 Exercise Set: 88

Answer

$6(p+q)(p-q)$

Work Step by Step

Factoring the $GCF= 6 $, the given expression, $ 6p^2-6q^2 ,$ is equivalent to \begin{array}{l} 6(p^2-q^2) .\end{array} Using $x^2-y^2=(x+y)(x-y)$ or the factoring of the difference of 2 squares, then the factored form of the expression, $ 6(p^2-q^2) ,$ is \begin{array}{l} 6(p+q)(p-q) .\end{array}
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