Answer
(a) $\left\{-i, i \right\}$
(b) $P(x) = (x-i)^2(x+i)^2$
Work Step by Step
$\bf{(a) \text{ Zeros}}$
Factor the polynomial completely to obtain:
$P(x) = (x^2+1)(x^2+1)
\\P(x) = [(x-i)(x+i)] \cdot [(x-i)(x+i)]$
Equate each unique factor to zero then solve each equation to obtain:
\begin{array}{ccc}
&x-i=0 &\text{or} &x+i=0
\\&x=i &\text{or} &x=-i\end{array}
Thus, the zeros of the function are: $\left\{-i, i \right\}$.
$\bf{(b) \text{ Completely Factored Form}}$
From part (a) above, the completely factored form of $P(x)$ is:
$P(x) = (x-i)(x+i)(x-i)(x+i)
\\P(x) = (x-i)^2(x+i)^2$
