Answer
$(a)\quad $Zeros: $\pm i $ , $\pm\sqrt{2}$
$(b)\quad P(x)=(x-\sqrt{2})(x+\sqrt{2})(x+i)(x-i)$
Work Step by Step
Complete Factorization Theorem (p.287)
$P$ factors into $n$ linear factors : $P(x)=a(x-c_{1})(x-c_{2})\cdots(x-c_{n})$
where $a$ is the leading coefficient of $P$ and $c_{1}, c_{1}, \ldots, c_{n}$ are the zeros of $P$.
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$(a)$
Temporarily substituting $t=x^{2},$
$x^{4}-x^{2}-2=0$
$t^{2}-t-2=0\quad $... factors of -2 that add up to -1 are -2 and +1
$(t-2)(t+1)=0\quad$... bring back x
$(x^{2}-2)(x^{2}+1)=0$
Either
$x^{2}-2=0\quad\Rightarrow\quad x=\pm\sqrt{2}$
or
$x^{2}+1=0\quad\Rightarrow\quad x=\pm i$
$(b)$
The leading coefficient of P is $a=1$, so, by the above theorem,
$P(x)=(x-\sqrt{2})(x+\sqrt{2})(x+i)(x-i)$
