Answer
$\{-1/3, 2, -4\} $
Work Step by Step
The problem, in fact, asks us to find roots of f.
First, let's divide
$(3x^{3}+7x^{2}-22x-8)\div(3x+1)\qquad (c=-1/3)$
$\begin{array}{lllllll}
\underline{-1/3}| & 3 & +7 & -22 & -8 & & \\
& & -1 & -2 & 8 & & \\
& -- & -- & -- & -- & -- & \\
& 3 & 6 & -24 & |\overline{ 0 } & &
\end{array}$
factor out 3 from the trinomial and multiplying with the divisor...
So, $(3x+1)$ is a factor of $f(x).$
$f(x)=(3x+1)(x^{2}+2x-8)$
we try to factor the trinomial by searching for two integers
whose product is $-8\cdot(1)=-8$ and whose sum is $2$....
We find $4$ and $-2:\quad $
$x^{2}+2x-8= x^{2}-2x+4x-8$
... factor in pairs ...
$=x(x-2)+4(x-2)$
$=(x-2)(x+4)$
So,
$f(x)=3(x+\displaystyle \frac{1}{3})(x-2)(x+4)$
and by (b) of the Factor Theorem,
the solution set of the equation $f(x)=0$
are the zeros of f:
$\{-1/3, 2, -4\} $
