Answer
zeros:$\displaystyle \qquad -\frac{1}{2},\ -1,\ \ -3\pm \sqrt{10}$
Work Step by Step
$P(x)=2x^{4}+15x^{3}+17x^{2}+3x-1,$
($1$ sign variations, we expect $1$ positive real zero)
$P(-x)=2x^{4}-15x^{3}+17x^{2}-3x-1,$
($3$ sign variations, we expect $3$ or $1$ negative real zeros)
$a_{0}=-1 \quad $p: $\pm 1,$
$a_{n}=2,\qquad $q: $\pm 1,\pm 2$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try$ \pm 1,\pm 1/2,$...
$\begin{array}{r|rrrrrrrrrrr}\hline
-1 & 2 & 15 & 17 & 3 & -1 & & \\
& & -2 & -13 & -4 & 1 & & \\
\hline & 2 & 13 & 4 & -1 & 0 & & \\
& & & & & & & \\
- 1/2 & 2 & 13 & 4 & -1 & & & \\
& & -1 & -6 & 1 & & & \\
\hline & 2 & 12 & -2 & 0 & & & \\
\end{array}$
$P(x)=(x+\displaystyle \frac{1}{2})(x+1)(2x^{2}+12x-2)=2(x+\frac{1}{2})(x+1)(x^{2}+6x-1)$
Find the zeros of the last factor with the quadratic formula
$x=\displaystyle \frac{-6\pm\sqrt{(6)^{2}-4(1)(-1)}}{2(1)}$
$=\dfrac{-6\pm\sqrt{40}}{2}=\dfrac{-6\pm 2\sqrt{10}}{2}=-3\pm \sqrt{10}$
zeros:$\displaystyle \qquad -\frac{1}{2},\ -1,\ \ -3\pm \sqrt{10}$