#### Answer

The roots of the polynomial equation are $x=1$ and $x=-2$.

#### Work Step by Step

Consider the polynomial equation ${{x}^{3}}-3x+2=0$.
Use the rational root theorem to observe:
${{a}_{0}}=2\text{ , }{{\text{a}}_{n}}=1$
The factors of ${{a}_{0}}$ are: 1,2 and the factors of ${{a}_{n}}$ are: 1
Possible roots are given by: $\pm \frac{1,2}{1}$.
$\frac{1}{1}$ is a root of the expression, so factor out $x-1$.
Compute $\frac{{{x}^{3}}-3x+2}{x-1}$ to get the rest of the equation ${{x}^{2}}+x-2$.
$\begin{align}
& \left( x-1 \right)\left( {{x}^{2}}+x-2 \right)=0 \\
& x=1,-2 \\
\end{align}$
Hence, the roots of the polynomial equation ${{x}^{3}}-3x+2=0$ are $x=1$ and $x=-2$.