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 4 - Polynomials - 4.6 Special Products - 4.6 Exercise Set: 108

Answer

$81a^4-1$

Work Step by Step

Use the formula $(a+b)(a-b)$ to multiply the last 2 factors to obtain: $=(9a^2+1)[(3a)^2-1^2] \\=(9a^2+1)(9a^2-1)$ Use the same formula again with $a=9a^2$ and $b=1$ to obtain: $=(9a^2)^2-1^2 \\=81a^4-1$
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