Answer
zeros:$\displaystyle \qquad 3,\ \frac{1\pm\sqrt{5}}{2}$
Work Step by Step
$P(x)=x^{4}-7x^{3}+14x^{2}-3x-9,$
($3$ sign variations, we expect $3$ or $1$ positive real zero)
$P(-x)=x^{4}+7x^{3}+14x^{2}+3x-9,$
(1 sign variations, we expect 1 negative real zeros)
$a_{0}=-9 \quad $p: $\pm 1,\pm 3,\pm 9$
$a_{n}=1,\qquad $q: $\pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try$ \pm$1, $\pm 3$...
$\begin{array}{r|rrrrrrrrrrr}\hline
3 & 1 & -7 & 14 & -3 & -9 & & \\
& & 3 & -12 & 6 & 9 & & \\
\hline & 1 & -4 & 2 & 3 & 0 & & \\
& & & & & & & \\
3 & 1 & -4 & 2 & 3 & & & \\
& & 3 & -3 & -3 & & & \\
\hline & 1 & -1 & -1 & 0 & & & \\
\end{array}$
$P(x)=(x-3)^{2}(x^{2}-x-1)$
Find the zeros of the second factor with the quadratic formula
$x=\displaystyle \frac{1\pm\sqrt{(1)^{2}-4(1)(-1)}}{2(1)}=\frac{1\pm\sqrt{5}}{2}$
zeros:$\displaystyle \qquad 3,\ \frac{1\pm\sqrt{5}}{2}$