Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Test - Page 402: 17


$(9y-5), (9y+5)$

Work Step by Step

Given: $81y^2-25$ Since, $a^2-b^2=(a-b)(a+b)$ This implies, $81y^2-25=(9y)^2-(5)^2$ we have $=(9y-5)(9y+5)$ Hence, factors are:$(9y-5), (9y+5)$
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