## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter P - Section P.5 - Factoring Polynomials - Exercise Set - Page 69: 82

#### Answer

$y(y^{2}+4)(y+2)(y-2)$

#### Work Step by Step

$y^{5}-16y=$ factor out the GCF, $y$ $=y(y^{4}-16)$ The parentheses hold a difference of squares, $=y[(y^{2})^{2}-4^{2})$ $=y(y^{2}+4)(y^{2}-4)$ The last parentheses hold a difference of squares, $=y(y^{2}+4)(y+2)(y-2)$

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