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: 110

Answer

$625a^4-50a^2+1$

Work Step by Step

RECALL: $x^my^m=(xy)^m$ Use the rule above to obtain: $=[(5a+1)(5a-1)]^2$ Multiply the two binomials using the formula $(a+b)(a-b)=a^2-b^2$ to obtain: $=[(5a)^2-1^2]^2 \\=(25a^2-1)^2$ Use the formula $(a-b)^2=a^2-2ab+b^2$ with $a=25a^2$ and $b=1$ to obtain: $=(25a^2)^2-2(25a^2)(1)+1^2 \\=625a^4-50a^2+1$
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