## Precalculus (6th Edition) Blitzer

$y(y^{2}+4)(y+2)(y-2)$
$y^{5}-16y=$ 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, $=y(y^{2}+4)(y+2)(y-2)$