Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies - Page 313: 16



Work Step by Step

Factoring the $GCF=9$, then the given expression, $ 9x^2-81 $, is \begin{array}{l} 9(x^2-9) .\end{array}Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of 2 squares, then the factored form of the expression, $ 9(x^2-9) $, is \begin{array}{l} 9(x+3)(x-3) .\end{array}
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