Answer
$x\displaystyle \in\left\{ 1-\sqrt 3, 1+\sqrt 3, -\frac{1}{3} \right\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f$ with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
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$f(x)=3x^{3}-5x^{2}-8x-2$
Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2$
$q:\qquad \pm 1, \pm3$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm \frac{1}{3}, \pm \frac{2}{3}$
Try for $x=-\frac{1}{3}:$
$\begin{array}{lllll}
\underline{-\frac{1}{3}}|& 3 & -5 & -8 & -2\\
& & -1& 2& 2\\
& -- & -- & -- & --\\
& 3 & -6& -6 & |\underline{0}
\end{array}$
$-\frac{1}{3}$ is a zero,
$f(x)=(x+\frac{1}{3})(3x^2-6x-6)$
Solving for the trinomial, using the quadratic formula for the quadratic equation of $ax^2+bx+c$, $x=\frac{-b\pm \sqrt {b^2-4ac}}{2a}$.
in this case, $3x^2-6x-6$, $x=\frac{6\pm \sqrt {(-6)^2-4 \times 3\times (-6)}}{2\times 3}=\frac{6\pm 6\sqrt {3}}{6}=1\pm\sqrt 3$.
Therefore, the real zeros of the equations are:
$x\displaystyle \in\left\{ 1-\sqrt 3, 1+\sqrt 3, -\frac{1}{3} \right\}$
