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: 109

Answer

$81t^4-72t^2+16$

Work Step by Step

RECALL: $x^my^m=(xy)^m$ Use the rule above to obtain: $=[(3t+2)(3t-2)]^2$ Multiply the two binomials using the formula $(a+b)(a-b)=a^2-b^2$ to obtain: $=[(3t)^2-2^2]^2 \\=(9t^2-4)^2$ Use the formula $(a-b)^2=a^2-2ab+b^2$ with $a=9t^2$ and $b=4$ to obtain: $=(9t^2)^2-2(9t^2)(4)+4^2 \\=81t^4-72t^2+16$
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