Answer
$x \in \{-i, i, 2 \}$
The zeros $-i$ and $i$ have multiplicity $2$.
Work Step by Step
$P(x)=x^5-2x^4+2x^3-4x^2+x-2$
Factorize the polynomial:
$x^4(x-2)+2x^2(x-2)+1(x-2)=(x-2)(x^4+2x^2+1)$,
Solve for the trinomial: let's let $x^2=k$.
$k^2+2k+1$
Factorize the trinomial $(k^2+2k+1)$,
(find two factors of $1$ whose sum is $2$:
($1$ and $1$))
$k^2+2k+1=k^2+k+k+1=(k+1)^2$.
Replace $x^2=k$:
$(x^2+1)^2$
Thus, the zeros are:
$(x-2)(x^2+1)^2=0$
$x \in \{-i, i, 2 \}$
The zeros $-i$ and $i$ have multiplicity $2$.
