Answer
$x\displaystyle \in\left\{-\frac{1}{2},2,4\right\}$
Work Step by Step
Factoring in pairs is not obvious here, so we try to find rational zeros of
$f(x)=2x^{3}-11x^{2}+10x+8$
Possible rational roots: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 2,\pm 4,\pm 8}{\pm 1,\pm 2}$
Testing with synthetic division, ... we try $x-2$
$\left.\begin{array}{l}
2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 &-11 &10 &8 \\\hline
&4 &-14 & -8 \\\hline
2& -7 &-4 & |\ \ 0 \end{array}$
$f(x)=(x-2)( 2x^{2}-7x-4)$
for the trinomial, we search for factors of $ac=-8$ whose sum is $b=-7.$
... we find -8 and +1.
$2x^{2}-7x-4=2x^{2}-8x+x-4$
$=2x(x-4)+(x-4)$
$=(x-4)(2x+1)$
$f(x)=(x-2)(x-4)(2x+1)=0$
when $x\displaystyle \in\left\{-\frac{1}{2},2,4\right\}$
