Answer
$3x^2y(y+5)(y-3)$
Work Step by Step
Factoring the $GCF=
3x^2y
$ results to $
3x^2y(y^2+2y-15)
$. The two numbers whose product is $
-15
$ and whose sum is $
2
$ are $\{
5,-3
\}.$ Using these numbers to decompose the middle term of the trinomial results to
\begin{align*}
3x^2y(y^2+2y-15)
\\\Rightarrow
3x^2y(y^2+5y-3y-15)
\\=
3x^2y[(y^2+5y)-(3y+15)]
\\=
3x^2y[y(y+5)-3(y+5)]
\\=
3x^2y[(y+5)(y-3)]
\\=
3x^2y(y+5)(y-3)
.\end{align*}
