Answer
a. $\displaystyle \pm 1, \pm 2,\pm\frac{1}{2}$
b. sample: $ 1$ is a zero...(see "step by step")
c. $-\displaystyle \frac{1}{2},1,2$
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, $a_{n}$.
------------------------
$f(x)=2x^{3}-5x^{2}+x+2$
a. candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 2$
$q:\qquad \pm 1, \pm 2$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm\frac{1}{2}$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| & 2 & -5 & 1 & 2\\
& & 2 & -3 & -2\\
& -- & -- & -- & --\\
& 2 & -3 & -2 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(2x^{2} -3x-2)$
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 $f(x)=0$
$(x-1)(2x+1)(x-2)=0$
$x\displaystyle \in\{-\frac{1}{2},1,2\}$