Answer
$y^{3}(3y-1)(5y+1)$
Work Step by Step
$ 15y^{5}-2y^{4}-y^{3}\qquad$...factor out the common term, $y^{3}$.
$=y^{3}(15y^{2}-2y-1)$
Two factors of $ac=-15$, whose sum is $b=-2$
are$ -5$ and $3$.
Rewrite the middle term and factor in pairs.
$y^{3}(15y^{2}-2y-1)$
$=y^{3}(15y^{2}-5y+3y-1)$
$=y^{3}[5y(3y-1)+(3y-1)]$
=$y^{3}(3y-1)(5y+1)$