Answer
$\left\{-\dfrac{1}{3}, -\dfrac{3}{2}, \dfrac{1}{2}\right\} $
Work Step by Step
The problem, in fact, asks us to find roots of f.
First, let's divide
$(12x^{3}+16x^{2}-5x-3)\div(2x+3)\qquad (c=-3/2)$
$\begin{array}{lllllll}
\underline{-3/2}| & 12 & +16 & -5 & -3 & & \\
& & -18 & 3 & 3 & & \\
& -- & -- & -- & -- & -- & \\
& 12 & -2 & -2 & |\overline{ 0 } & &
\end{array}$
factor out 2 from the trinomial and multiplying with the divisor...
So, $(2x+3)$ is a factor of $f(x).$
$f(x)=(2x+3)(6x^{2}-x-1)$
we try to factor the trinomial by searching for two integers
whose product is $-1\cdot(6)=-6$ and whose sum is $-1$....
We find $2$ and $-3:\quad $
$6x^{2}-x-1= 6x^{2}-3x+2x-1$
... factor in pairs ...
$=3x(2x-1)+1(2x-1)$
$=(2x-1)(3x+1)$
... factor out 2 from the first parentheses and 3 from the second parentheses ...
$=2(x-\displaystyle \frac{1}{2})3(x+1/3)$
So,
$f(x)=12(x-\displaystyle \frac{1}{2})(x+1/3)(x+3/2)$
and by (b) of the Factor Theorem,
the solution set of the equation $f(x)=0$
are the zeros of f:
$\left\{-\dfrac{1}{3}, -\dfrac{3}{2}, \dfrac{1}{2}\right\} $
