Answer
$x\displaystyle \in\left\{-10,10,-\frac{5}{2}\right\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ 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, $a_{n}$.
------------------------
$2x^{3}+5x^{2}-200x-500=0$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm4, \pm5, \pm 10, \pm20, \pm25, \pm50, \pm100, \pm 125, \pm250, \pm500$
$q:\qquad \pm 1, \pm 2$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm4, \pm5, \pm 10, \pm20, \pm25, \pm50, \pm100, \pm 125, \pm250, \pm500, \pm\frac{1}{2}, \pm\frac{5}{2}, \pm\frac{25}{2}, \pm\frac{125}{2}$
b. Try for $x=10:$
$\begin{array}{lllll}
\underline{10}| & 2 & 5 & -200 & -500\\
& & 20 & 250 & 500\\
& -- & -- & -- & --\\
& 2 & 25 & 50 & |\underline{0}
\end{array}$
$10$ is a zero,
$(x-10)(2x^{2}+25x +50)=0$
c. Factorize the trinomial factor $(2x^{2} +25x+50)$
(find two factors of $50(2)=100$ whose sum is $25):$
$(5$ and $20$)
$2x^{2} +25x+50=2x^{2} +5x+20x+50 \quad$...factor in pairs ...
$x(2x+5)+10(2x+5)=(2x+5)(x+10)$
$(x-10)(2x+5)(x+10)=0$
$x\displaystyle \in\left\{-10,10,-\frac{5}{2}\right\}$
