Answer
$P(x)=\dfrac{1}{8}(2x+3)^2(x-2)^2(x^2+3x+9)^2$
Zeros: $-\dfrac{3}{2}$ and $2$
Refer to the graph below.
Work Step by Step
Factor the polynomial completely to obtain:
\begin{align*}
P(x)&=\frac{1}{8}\left[(2x^4+3x^3)+(-16x-24)\right]^2\\
P(x)&=\frac{1}{8}\left[x^3(2x+3)+(-8)(2x+3)\right]^2\\
P(x)&=\frac{1}{8}\left[(2x+3)(x^3-8)\right]^2\\
P(x)&=\frac{1}{8}\left[(2x+3)(x^3-8)\right]^2\\
P(x)&=\frac{1}{8}\left[(2x+3)(x-2)(x^2+3x+9)\right]^2\\
P(x)&=\frac{1}{8}(2x+3)^2(x-2)^2(x^2+3x+9)^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*}
(2x+3)^2&=0 &\text{or}& &(x-2)^2=0& &\text{or}& &(x^2+3x+9)^2=0\\
2x+3&=0 &\text{or}& &x-2=0& &\text{or}& &x^2+3x+9=0\\
2x&=-3 &\text{or}& &x=2& &\text{or}& &\text{ (no real solution)}\\
x&=-\frac{3}{2}
\end{align*}
Both real roots have multiplicity $2$.
Use a graphing utility to graph $P(x)$.
Refer to the graph above.
