#### Answer

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

#### Work Step by Step

The problem, in fact, asks us to find zeros of f.
First, let's divide
$(2x^{3}-3x^{2}-11x+6)\div(x+2)\qquad (c=-2)$
$\begin{array}{lllllll}
\underline{-2}| & 2 & -3 & -11 & 6 & & \\
& & -4 & 14 & -6 & & \\
& -- & -- & -- & -- & -- & \\
& 2 & -7 & 3 & |\overline{ 0 } & &
\end{array}$
So, $(x+2)$ is a factor of $f(x).$
$f(x)=(x+2)(2x^{2}-7x+3)$
we try to factor the trinomial by searching for two integers
whose product is $2\cdot(3)=6$ and whose sum is $-7$....
We find $-1$ and $-6:\quad $
$2x^{2}-7x+3= 2x^{2}-x-6x+3$
... factor in pairs ...
$=x(2x-1)-3(2x-1)$
$=(2x-1)(x-3)$
... factor out 2 from the first parentheses ...
$=2(x-\displaystyle \frac{1}{2})(x-3)$
So,
$f(x)=2(x-\displaystyle \frac{1}{2})(x-3)(x+2)$
and by (b) of the Factor Theorem,
the solution set of the equation $f(x)=0$
are the zeros of f:
$\{-2,\ \displaystyle \frac{1}{2},\ 3\} $