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