## Intermediate Algebra (6th Edition)

$3x^2y(y+5)(y-3)$
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*}