#### Answer

solution set: $\{ -3,\ \displaystyle \frac{1}{2},\ 2 \}$

#### Work Step by Step

The graph suggests that x=$-3$ is a solution to the equation.
We check by synthetic division(the last number in the last row is $f(-3)$
$( 2x^{3}+x^{2}-13x+6)\div(x+3)$
$\begin{array}{lllllll}
\underline{-3}| & 2 & 1 & -13 & 6 & & \\
& & -6 & 15 & -6 & & \\
& -- & -- & -- & -- & -- & \\
& 2 & -5 & 2 & |\overline{ 0 } & &
\end{array}$
By the Factor Theorem, $-3$ is a zero, $(x+3)$ is a factor of $f(x)$.
$f(x)=(x+3)(2x^{2}-5x+2)$
we try to factor the trinomial by searching for two integers
whose product is $2\cdot 2=4$ and whose sum is $-5$....
We find $-1$ and $-4:\quad $
$ 2x^{2}-5x+2=2x^{2}-4x-x+2\quad $...factor in pairs
$=2x(x-2)-(x-2) = (2x-1)(x-2)$
... factor out 2 from the first...
$=2(x-\displaystyle \frac{1}{2})(x-2)$
So, f(x) has three factors of form $(x-c)$,
$f(x)=2(x-\displaystyle \frac{1}{2})(x-2)(x+3)$
and by (b) of the Factor Theorem,
zeros of f: $ \displaystyle \frac{1}{2},\ 2,\ -3$
make up the solution set of the equation $f(x)=0:$
$\{ -3,\ \displaystyle \frac{1}{2},\ 2 \}$