Answer
$y=-3$ or $y=1$ or $y=0$
Work Step by Step
$ y^{3}+2y^{2}-3y=0\qquad$...factor out the common term, $y$.
$y(y^{2}+2y-3)=0$
... Searching for two factors of $ac=-3$ whose sum is $b=2,$
we find$\qquad-1$ and $3.$
Rewrite the middle term and factor in pairs:
$y(y^{2}-y+3y-3)=0$
$y[y(y-1)+3(y-1)]=0$
$ y(y-1)(y+3)=0\qquad$...apply the principle of zero products.
$y=0$
$y-1=0$
$y=1$
$y+3=0$
$y=-3$
Type the equation into a graphing utility and see
that the graph intercepts the x-axis at $-3,1$ and $0$.