Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.5 - Factoring Polynomials - Exercise Set - Page 69: 81

Answer

$y(y^{2}+9)(y+3)(y-3)$

Work Step by Step

$y^{5}-81y=$ Factor out the GCF, $y$ $=y(y^{4}-81)$ The parentheses hold a difference of squares, $=y[(y^{2})^{2}-9^{2})$ $=y(y^{2}+9)(y^{2}-9)$ The last parentheses hold a difference of squares, $=y(y^{2}+9)(y+3)(y-3)$
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