Answer
a. zeros: $-2,2,3$
b. see image
Work Step by Step
$P(x)= x^{3}-3x^{2}-4x+12,$
($4$ 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}=12 \quad $p: $\pm 1,\pm2,\pm 3,\pm 4 \pm6,\pm12$
$a_{n}=1,\qquad $q: $\pm 1$
Possible rational zeros $\displaystyle \frac{p}{q}:\quad$
with synthetic division, we try$ \pm 1,\pm 2,\pm 3$...
$\begin{array}{r|rrrrrrrrrrr}\hline
2 & 1 & -3 & -4 & 12 & & & \\
& & 2 & -2 & -12 & & & \\
\hline & 1 & -1 & -6 & 0 & & & \\
& & \swarrow & & & & & \\
-2 & 1 & -1 & -6 & & & & \\
& & -2 & 6 & & & & \\
\hline & 1 & -3 & 0 & & & & \\
\end{array}$
$P(x)=(x-2)(x+2)(x-3)$
$P(0)=12\qquad $(y intercept)
Calculate $P(-0.5)=13.25$ and $P(0.5)=9.375 $
to figure whether the graph turns before or after the y-intercept.
(Here, the graph falls through the y-intercept.)
The leading coefficient is positive, so the left far end is pointing down.\\\\
From left to right, the graph comes from far below x,
crosses x at (-2,0), rises up, turns,
crosses y at (0,12), falls to the next zero at (2,0),
falls below x, turns, and crosses x again at (3,0),
from where it continues rising.