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 - 5.6 Factoring: A General Strategy - 5.6 Exercise Set: 80

Answer

$a(b^2+9a^2)(b+3a)(b-3a)$

Work Step by Step

Factoring the $GCF= a $, then the given expression, $ b^4a-81a^5 $ is equivalent to \begin{array}{l} a(b^4-81a^4) .\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 given expression, $ a(b^4-81a^4) $, is \begin{array}{l} a(b^2+9a^2)(b^2-9a^2) \\\\= a(b^2+9a^2)(b+3a)(b-3a) .\end{array}
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