Answer
a. $\displaystyle \pm 1, \pm 2,\pm 3,\pm 6,\pm\frac{1}{2}, \pm\frac{3}{2}$
b. sample: $ 3$ is a zero...(see "step by step")
c. $-2,\displaystyle \frac{1}{2},3$
Work Step by Step
see The Rational Zero Theorem:
... If $\displaystyle \frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients,
then $p$ is a factor of the constant term, $a_{0}$, and
$q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$f(x)=2x^{3}-3x^{2}-11x+6$
a. candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 2,\pm 3,\pm 6,$
$q:\qquad \pm 1, \pm 2$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm 3,\pm 6,\pm\frac{1}{2}, \pm\frac{3}{2}$
b. Try for $x=3:$
$\begin{array}{lllll}
\underline{3}| & 2 & -3 & -11 & 6\\
& & 6 & 9 & -6\\
& -- & -- & -- & --\\
& 2 & 3 & -2 & |\underline{0}
\end{array}$
$3$ is a zero,
$f(x)=(x-3)(2x^{2} +3x-2)$
c. Factorize the trinomial factor $(2x^{2} +3x-2)$
(find two factors of $2(-2)=-4$ whose sum is $3):$
$($4 and $-1$)
$2x^{2} +3x-2=2x^{2} +4x-x-2 \quad$...factor in pairs ...
$=2x(x+2)-(x+2)=(2x-1)(x+2)$
$f(x)=(x-1)(2x-1)(x+2)$
The zeros of f satisfy $f(x)=0$
$(x-3)(2x-1)(x+2)=0$
$x\displaystyle \in\{-2,\frac{1}{2},3\}$
