Answer
$x\displaystyle \in\{ 1 - i, 1 + i, -1\}$
The zero $-1$ has multiplicity $2$.
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f$ with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$f(x)=x^4-x^2+2x+2$
a. candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad\pm 1, \pm 2$
b. Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}| & 1 & 0 & -1 & 2& 2\\
& & -1 & 1 & 0& -2\\
& -- & -- & -- & --\\
& 1 & -1 & 0& 2 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)(x^3-x^{2} +2)$
Try for $x=-1$.
$\begin{array}{lllll}
\underline{-1}| & 1 & -1 & 0& 2\\
& & -1 & 2& -2\\
& -- & -- & -- & --\\
& 1 & -2& 2 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)^2(x^2-2x+2)$,
c. Solve for the trinomial using the quadratic formula for the quadratic equation of $ax^2+bx+c$, $x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$
in this case, $x^2-2x+2$, $x=\frac{2 \pm \sqrt {(-2)^2-4 \times1 \times 2}}{2 \times 1}=\frac{2\pm2i}{2}=1\pm i$
$x\displaystyle \in\{ 1 - i, 1 + i, -1\}$
The zero $-1$ has multiplicity $2$.
