Answer
$P(x)=(x+3)(x-2)(x+2)$
Zeros: $-3, -2\text{ and } 2$
Refer to the graph below.
Work Step by Step
Factor the polynomial completely to obtain:
\begin{align*}
P(x)&=(x^3+3x^2)+(-4x-12)\\
&=x^2(x+3)+(-4)(x+3)\\
&=(x+3)(x^2-4)\\
&=(x+3)(x-2)(x+2)\\
\end{align*}
To find the zeros, use the Zero-Product Property by equating each factor to $0$, then solve each equation to obtain:
\begin{align*}
x+3&=0 &\text{or}& &x-2=0& &\text{or}& =&x+2=0\\
x&=-3 &\text{or}& &x=2& &\text{or}& =&x=-2\\
\end{align*}
Use a graphing utility to graph $P(x)$.
Refer to the graph above.
