Answer
$x\in\{-1,2\}$
Work Step by Step
We are searching for the real zeros of $f(x)=x^{4}-x^{3}+2x^{2}-4x-8$
Possible rational zeros: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 2,\pm 4,\pm 8}{\pm 1}$
Testing with synthetic division, try $x+1$
$\left.\begin{array}{l}
-1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 &-1 &2 &-4 &-8 \\\hline
&-1 &2 &-4 & 8 \\\hline
1& -2 & 4 &-8 & |\ \ 0 \end{array}$
$f(x)=(x+1)( x^{3}-2x^{2}+4x-8)$
factor in pairs:
$x^{3}-2x^{2}+4x-8=x^{2}(x-2)+2(x-2)$
$=(x-2)(x^{2}+2)$
$x^{2}+2$ has no real zeros and can not be factored any further.
$f(x)=(x+1)(x-2)(x^{2}+2)$
$f(x) =0$ when $x\in\{-1,2\}$
