Answer
(a) $\left\{-i, i \sqrt2, -\sqrt2 \right\}$.
(b) $P(x) = (x-\sqrt2)(x+\sqrt2)(x-i)(x+i)$
Work Step by Step
$\bf{(a) \text{ Zeros}}$
Factor the polynomial completely to obtain:
$P(x) = (x^2-2)(x^2+1)
\\P(x) = (x-\sqrt2)(x+\sqrt2)(x-i)(x+i)$
Equate each unique factor to zero then solve each equation to obtain:
\begin{array}{cccccc}
&x-\sqrt2=0 &\text{or} &x+\sqrt2=0 &\text{or} &x-i=0 &\text{or} &x+i=0
\\&x=\sqrt2 &\text{or} &x=-\sqrt2 &\text{or} &x=i &\text{or} &x=-i
\end{array}
Thus, the zeros of the function are: $\left\{-i, i \sqrt2, -\sqrt2 \right\}$.
$\bf{(b) \text{ Completely Factored Form}}$
From part (a) above, the completely factored form of $P(x)$ is:
$P(x) = (x-\sqrt2)(x+\sqrt2)(x-i)(x+i)$
