Answer
$P(x)=(x-2)^2(x^2+2x+4)$
Zero: $2 \text{ (multiplicity 2)}$
Refer to the graph below.
Work Step by Step
Factor the polynomial completely to obtain:
\begin{align*} P(x)&=(x^4-2x^3)+(-8x+16)\\
&=x^3(x-2)+(-8)(x-2)\\
&=(x-2)(x^3-8)\\
&=(x-2)(x-2)(x^2+2x+4)\\
&=(x-2)^2(x^2+2x+4)\\
\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-2)^2&=0 &\text{or}& &x^2+2x+4=0\\
x-2&=0 &\text{or}& &\text{(no real solution)} \\
x&=2 \text{ (multiplicity 2)}\\
\end{align*}
The zero of the function is $2$.
Use a graphing utility to graph $P(x)$.
Refer to the graph above.
