Answer
$2y^{2}(y-4)(y^{2}+4y+16)$
Work Step by Step
$ 2y^{5}-128y^{2}\qquad$...factor out the common term, $2y^{2}$
$=2y^{2}(y^{3}-64)\qquad$...use the difference of two cubes: $\mathrm{A}^{3}-\mathrm{B}^{3} = (\mathrm{A}-\mathrm{B})(\mathrm{A}^{2} + \mathrm{A}\mathrm{B} + B^{2})$
$=2y^{2}(y^{3}-4^{3})$
$=2y^{2}(y-4)(y^{2}+4y+16)$
