Answer
$x\in\{-2i, -\sqrt 3i, \sqrt 3i, 2i, 1\}$
Work Step by Step
$P(x)=x^5-x^4+7x^3-7x^2+12x-12$,
Factor the polynomial:
$x^4(x-1)+7x^2(x-1)+12(x-1)=(x-1)(x^4+7x^2+12)$.
Solve for the trinomial:
Let $x^2=k$.
$x^4+7x^2+12=k^2+7k+12$
To solve for the quadratic $k^2+7k+12$, factor out in a way such that we find two factors of $12$ whose sum is $7$:
($+3$ and $+4$).
$k^2+7k+12=k^2+3k+4k+12=k(k+3)+4(k+3)=(k+3)(k+4)$,
Since $k=x^2$, $(x^2+3)(x^2+4)=0$. $x=\pm2i$, $x=\pm \sqrt 3i$
$x \in \{-2i, -\sqrt 3i, \sqrt 3i, 2i, 1\}$
