Answer
$x\in \{ -\frac{1}{3},1,3 \}$
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, $a_{n}$.
------------------------
$3x^3+2x-1=8x^2-4,$
$3x^3-8x^2+2x+3=0$
a. candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm3,$
$q:\qquad \pm 1,\pm3$
$\displaystyle \frac{p}{q}:\qquad \pm 1,\pm3,\pm\frac{1}{3},$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| &3& -8 & 2 & 3\\
& & 3&-5 & -3\\
& -- & -- & -- & --\\
& 3&-5 & -3 & |\underline{0}
\end{array}$
$f(x)=(x-1)(3x^2-5x-3),$
$(3x^2+x-6x-3),$
$x(3x+1)-3(3x+1),$
$(3x+1)(x-3),$
$f(x)=(x-1)(3x+1)(x-3))$
$x\in \{ -\frac{1}{3},1,3 \}$
