Answer
zeros:$\displaystyle \qquad \frac{1}{2},\ 3,\ \ \frac{-2\pm\sqrt{6}}{2}$
Work Step by Step
$P(x)=4x^{5}-18x^{4}-6x^{3}+91x^{2}-60x+9,$
($4$ sign variations, we expect $4,2$ or $0$ positive real zeros)
$P(-x)=-4x^{5}-18x^{4}+6x^{3}+91x^{2}+60x+9,$
($1$ sign variations, we expect $1$ negative real zeros)
$a_{0}=9 \quad $p: $\pm 1,\pm 3,\pm 9$
$a_{n}=4,\qquad $q: $\pm 1,\pm 2,\pm 4$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try $ \pm 1,\pm 3,\pm 1/2$...
$\begin{array}{r|rrrrrrrrrrr}\hline
3 & 4 & -18 & -6 & 91 & -60 & 9 & \\
& & 12 & -18 & -72 & 57 & -9 & \\
\hline & 4 & -6 & -24 & 19 & -3 & 0 & \\
& & & \swarrow & & & & \\
3 & 4 & -6 & -24 & 19 & -3 & & \\
& & 12 & 18 & -18 & 3 & & \\
\hline & 4 & 6 & -6 & 1 & 0 & & \\
& & & \swarrow & & & & \\
1/2 & 4 & 6 & -6 & 1 & & & \\
& & 2 & 4 & -1 & & & \\
\hline & 4 & 8 & -2 & 0 & & & \\
\end{array}$
$P(x)=(x-3)^{2}(x-\displaystyle \frac{1}{2})(4x^{2}+8x-2)$
$=2(x-3)^{2}(x-\displaystyle \frac{1}{2})(2x^{2}+4x-1)$
Find the zeros of the last factor with the quadratic formula
$x=\displaystyle \frac{-4\pm\sqrt{(-4)^{2}-4(2)(-1)}}{2(2)}=\frac{-4\pm\sqrt{24}}{4}=\frac{-4\pm 2\sqrt{6}}{4}=\frac{-2\pm\sqrt{6}}{2}$
zeros:$\displaystyle \qquad \frac{1}{2},\ 3,\ \ \frac{-2\pm\sqrt{6}}{2}$
