Answer
zeros:$\displaystyle \qquad -1,\ 3, \ \ \frac{-1\pm\sqrt{61}}{6}$
Work Step by Step
$P(x)=3x^{4}-5x^{3}-16x^{2}+7x+15,$
($2$ sign variations, we expect $2$ or $0$ real zero)
$P(-x)=3x^{4}+5x^{3}-16x^{2}-7x+15$
($2$ sign variations, we expect $2$ or $0$ negative real zeros)
$a_{0}=15 \quad $p: $\pm 1,\pm 3,\pm 5,\pm 15$
$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,...
$\begin{array}{r|rrrrrrrrrrr}\hline
-1 & 3 & -5 & -16 & 7 & 15 & & \\
& & -3 & 8 & 8 & -15 & & \\
\hline & 3 & -8 & -8 & 15 & 0 & & \\
& & & \swarrow & & & & \\
3 & 3 & -8 & -8 & 15 & & & \\
& & 9 & 3 & -15 & & & \\
\hline & 3 & 1 & -5 & 0 & & & \\
\end{array}$
$P(x)=(x+1)(x-3)(3x^{2}+x-5)$
Find the zeros of the second factor with the quadratic formula
$x=\displaystyle \frac{-1\pm\sqrt{1^{2}-4(3)(-5)}}{2(3)}=\frac{-1\pm\sqrt{61}}{6}$
zeros:$\displaystyle \qquad -1,\ 3, \ \ \frac{-1\pm\sqrt{61}}{6}$
