Answer
$-4y(y-2)(y-5)$
Work Step by Step
$-4y^{3}+28y^{2}-40y\qquad$...factor out the common term, $-4y$
$=-4y(y^{2}-7y+10)$
Two factors of $ac=10$, whose sum is $b=-7$
are $-5$ and $-2$.
Rewrite the middle term and factor in pairs.
$-4y(y^{2}-7y+10)$
$=-4y(y^{2}-2y-5y+10)$
$=-4y[y(y-2)-5(y-2)]\qquad$...factor out the common binomial factor,$y-2$.
$=-4y(y-2)(y-5)$
