Answer
zeros: $-1, \displaystyle \ \ \frac{-1\pm\sqrt{13}}{3}$
Work Step by Step
$P(x)=3x^{3}+5x^{2}-2x-4,$
($1$ sign variations, we expect $1$ real zero)
$P(-x)=-3x^{3}+5x^{2}+2x-4,$
($3$ sign variations, we expect $3$ or $1$ negative real zeros)
$a_{0}=-4 \quad $p: $\pm 1,\pm 2,\pm 4$
$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,...
$\begin{array}{r|rrrrrrrrrrr}\hline
-1 & 3 & 5 & -2 & -4 & & & \\
& & -3 & -2 & 4 & & & \\
\hline & 3 & 2 & -4 & 0 & & & \\
& & & & & & & \\
\end{array}$
$P(x)=(x+1)(3x^{2}+2x-4)$
Find the zeros of the second factor with the quadratic formula
$x=\displaystyle \frac{-2\pm\sqrt{2^{2}-4(3)(-4)}}{2(3)}=\frac{-2\pm\sqrt{52}}{6}=\frac{-2\pm 2\sqrt{13}}{6}=\frac{-1\pm\sqrt{13}}{3}$
zeros: $-1, \displaystyle \ \ \frac{-1\pm\sqrt{13}}{3}$