Answer
zeros:$\displaystyle \qquad -\frac{1}{3},\ \ \ 1\pm\sqrt{3}$
Work Step by Step
$P(x)=3x^{3}-5x^{2}-8x-2,$
($1$ sign variations, we expect $1$ positive real zero)
$P(-x)=-3x^{3}-5x^{2}+8x-2,$
($2$ sign variations, we expect $2$ or $0$ negative real zeros)
$a_{0}=-2 \quad $p: $\pm 1,\pm 2$
$a_{n}=3,\qquad $q: $\pm 1,\pm 3$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try$ \pm 1,\pm 2, \pm 1/3$...
$\begin{array}{r|rrrrrrrrrrr}\hline
- 1/3 & 3 & -5 & -8 & -2 & & & \\
& & -1 & 2 & 2 & & & \\
\hline & 3 & -6 & -6 & 0 & & & \\
\end{array}$
$P(x)=(x+\displaystyle \frac{1}{3})(3x^{2}-6x-6)=3(x+\frac{1}{3})(x^{2}-2x-2)$
Find the zeros of the last factor with the quadratic formula
$x=\displaystyle \frac{2\pm\sqrt{(-2)^{2}-4(1)(-2)}}{2(1)}=\frac{2\pm\sqrt{12}}{2}=\frac{2\pm 2\sqrt{3}}{2}=1\pm\sqrt{3}$
zeros:$\displaystyle \qquad -\frac{1}{3},\ \ \ 1\pm\sqrt{3}$