Answer
$y^{3}(2y-3)(5y-1)$
Work Step by Step
$ 10y^{5}-17y^{4}+3y^{3}\qquad$...factor out the common term, $y^{3}$.
$=y^{3}(10y^{2}-17y+3)$
Two factors of $ac=30$, whose sum is $b=-17$
are$ -15$ and $-2$.
Rewrite the middle term and factor in pairs.
$y^{3}(10y^{2}-17y+3)$
$=y^{3}(10y^{2}-15y-2y+3)$
$=y^{3}[5y(2y-3)-(2y-3)]$
=$y^{3}(2y-3)(5y-1)$