## College Algebra 7th Edition

$P(x)=(x+1)^2(x-1)$ Zeros: $1$ and $-1$ (multiplicity $2$ Refer to the graph below.
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.