Answer
$P(x)=(x+1)(x-2)^2(x+2)(x-3)$
zeros: $\quad -2, -1, 2, 3$
Work Step by Step
$P(x)=x^{5}-4x^{4}-3x^{3}+22x^{2}-4x-24,$
($3$ sign variations, we expect $3$ or $1$ real zeros)
$P(-x)=-x^{5}-4x^{4}+3x^{3}+22x^{2}+4x-24,$
($2$ sign variations, we expect $2 $or $0$ negative real zero)
$a_{0}=-24 \quad $p: $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12,\pm 24$
$a_{n}=1,\qquad $q: $\pm 1,\pm 2,\pm 4$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try$ \pm$1, $\pm$2,...
$\begin{array}{r|rrrrrrrrrrr}\hline
-1 & 1 & -4 & -3 & 22 & -4 & -24 & \\
& & -1 & 5 & -2 & -20 & 24 & \\
\hline & 1 & -5 & 2 & 20 & -24 & 0 & \\
& & & \swarrow & & & & \\
2 & 1 & -5 & 2 & 20 & -24 & & \\
& & 2 & -6 & -8 & 24 & & \\
\hline & 1 & -3 & -4 & 12 & 0 & & \\
& & \swarrow & & & & \\
2 & 1 & -3 & -4 & 12 & & & \\
& & 2 & -2 & -12 & & & \\
\hline & 1 & -1 & -6 & 0 & & & \\
& & \swarrow & & & & \\
-2 & 1 & -1 & -6 & & & & \\
& & -2 & 6 & & & & \\
\hline & 1 & -3 & 0 & & & & \\
& & & & & & & \\
3 & 1 & -3 & & & & & \\
& & 3 & & & & & \\
\hline & 1 & 0 & & & & &
\end{array}$
$P(x)=(x+1)(x-2)(x-2)(x+2)(x-3)$
zeros: $\quad -2, -1, 2, 3$
