Answer
The zeros of the function are $-3,\frac{1}{2}\text{ and }2$
Work Step by Step
Let us consider the function $f\left( x \right)=2{{x}^{3}}+{{x}^{2}}-13x+6$
It can be clearly seen that one of the roots of the function is 2. Therefore, $\left( x-2 \right)$ is a factor of the function.
We divide the function with $\left( x-2 \right)$ to get other factors
$\frac{2{{x}^{3}}+{{x}^{2}}-13x+6}{x-2}=2{{x}^{2}}+5x-3$
So, the function can be written as given below:
$\begin{align}
& f\left( x \right)=\left( 2{{x}^{2}}+5x-3 \right)\left( x-2 \right) \\
& =\left( x+3 \right)\left( x-2 \right)\left( x-\frac{1}{2} \right)
\end{align}$
Thus, the zeros of the equation are $-3,\frac{1}{2}\text{ and }2$.
