Answer
$x\displaystyle \in\{-1, -2i, 2i, 2 \}$
Work Step by Step
See The Rational Zero Theorem:
If $\displaystyle \frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
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$f(x)=x^4-x^{3}+2x^{2}-4x-8$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm 4, \pm 8$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm 4, \pm 8$
b. Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}| & 1 & -1 & 2 & -4 & -8\\
& & -1 & 2 & -4& 8\\
& -- & -- & -- & --\\
& 1 & -2 & 4 & -8 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)(x^3-2x^{2} +4x-8)$,
To solve for quadrinomial,
Try for $x=2:$
$\begin{array}{lllll}
\underline{2}| & 1 & -2 & 4 & -8\\
& & 2 & 0 & 8\\
& -- & -- & -- & --\\
& 1 & 0 & 4 & |\underline{0}
\end{array}$
$2$ is azero,
$f(x)=(x+1)(x-2)(x^2+4)$
c. To solve for the quadratic,
set the quadratic equal to $0$ and solve.
$(x^2+4)=0$,
$x^2=-4$
$x= \pm 2i$
The zeros of f satisfy $f(x)=0$
$(x+1)(x-2)(x -2i)(x+2i)=0$
$x\displaystyle \in\{-1, -2i, 2i, 2 \}$
