#### Answer

(a) $\left\{-2i, 2i, -2, 2\right\}$.
(b) $P(x) = (x-2)(x+2)(x-2i)(x+2i)$

#### Work Step by Step

$\bf{(a) \text{ Zeros}}$
Factor the polynomial completely to obtain:
$P(x) = (x^2-4)(x^2+4)
\\P(x) = ((x-2)(x+2)(x^2+4)
\\P(x) = (x-2)(x+2)(x-2i)(x+2i)$
Equate each unique factor to zero then solve each equation to obtain:
\begin{array}{cccccc}
&x-2=0 &\text{or} &x+2=0 &\text{or} &x-2i=0 &\text{or} &x+2i=0
\\&x=2 &\text{or} &x=-2 &\text{or} &x=2i &\text{or} &x=-2i
\end{array}
Thus, the zeros of the function are: $\left\{-2i, 2i, -2, 2\right\}$.
$\bf{(b) \text{ Completely Factored Form}}$
From part (a) above, the completely factored form of $P(x)$ is:
$P(x) = (x-2)(x+2)(x-2i)(x+2i)$