Answer
$P(x)=(x+1)^2(x-1)$
Zeros: $1$ and $-1$ (multiplicity $2$
Refer to the graph below.
Work Step by Step
Factor the polynomial completely to obtain:
\begin{align*}
P(x)&=(x^3+x^2)+(-x-1)\\
&=x^2(x+1)+(-1)(x+1)\\
&=(x+1)(x^2-1)\\
&=(x+1)(x-1)(x+1)\\
&=(x+1)^2(x-1)
\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+1)^2&=0 &\text{or}& &x-1=0\\
x+1&=\pm0 &\text{or}& &x=1\\
x&=-1 \text{ (multiplicity 2)} &\text{or}& &x=1\end{align*}
Use a graphing utility to graph $P(x)$.
Refer to the graph above.
