Answer
$x\displaystyle \in\left\{-4,\ -\frac{1}{2},\ 2\right\}$
Work Step by Step
We try to find rational zeros of $f(x)=2x^{4}+x^{3}-24x^{2}+20x+16$
There are at most 4 real zeros.
Possible rational zeros: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 2,\pm 4,\pm 8,\pm 16}{\pm 1,\pm 2}$
Testing with synthetic division, ... we eventually try $x-2$
$\left.\begin{array}{l}
2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & 1 & -24 & 20 &16 \\\hline
& 4 & 10 & -28 & -16 \\\hline
2& 5 & -14 &-8 & |\ \ 0 \end{array}$
$f(x)=(x-2)(2x^{3}+5x^{2}-14x-8)$
Testing with synthetic division, ... again, try $x-2$
$\left.\begin{array}{l}
2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & 5 & -14 &-8 \\\hline
& 4 & 18 & +8 \\\hline
2& 9 &4 & |\ \ 0 \end{array}$
$f(x)=(x-2)^{2}(2x^{2}+9x+4)$
Testing with synthetic division, ... try $x+4$
$\left.\begin{array}{l}
-4 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & 9 & 4 \\\hline
& -8 & -4 \\\hline
2 &1 & |\ \ 0 \end{array}$
$f(x)=(x-2)^{2}(x+4)(2x+1)=(x-2)^{2}(x+4)\displaystyle \cdot 2(x+\frac{1}{2})$
$f(x)=(x-2)^{2}(x+4)\displaystyle \cdot 2(x+\frac{1}{2})=0 $
when $x\displaystyle \in\left\{-4,\ -\frac{1}{2},\ 2\right\}$
