Answer
$(a)\quad $Zeros: $0$ and $1\pm i$
$(b)\quad P(x)=x(x-1-i)(x-1+i)$
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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$(a)\quad$
$x^{3}-2x^{2}+x=0$
$x(x^{2}-2x+2)=0$
So,
either
$x=0\Rightarrow\quad x=0$ is a zero
or
$x^{2}-2x+2=0\displaystyle \Rightarrow\quad x=\frac{2\pm\sqrt{4-8}}{2}=\frac{2(1\pm i)}{2}=1\pm i$
Zeros: $0$ and $1\pm i$
$(b)$
The leading coefficient of P is $a=1$, so, by the above theorem,
$P(x)=x(x-1-i)(x-1+i)$
