Answer
$(a)\quad $Zeros: $\pm i $
$(b)\quad P(x)=(x-i)^{2}(x+i)^{2}$
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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$(a)\quad$
$x^{4}+2x^{2}+1=0$ ... recognize a square of a binomial
$(x^{2}+1)^{2}=0$
So,
$x^{2}+1=0\Rightarrow\quad x=\pm i $
Zeros: $\pm i $
$(b)$
The leading coefficient of P is $a=1$, so, by the above theorem,
$P(x)=(x^{2}+1)^{2}$
$=[(x-i)(x+i)]^{2}$
$=(x-i)^{2}(x+i)^{2}$
