Answer
$P(x)=-x(x+3)(x-4)$
Zeros: $-3, 0,4,$ all single.
Work Step by Step
Factor out $-x$
$P(x)=-x(x^{2}-x-12)\\\\$
the quadratic can be factored by finding two factors of ($-12$) whose sum is $-1$.
These are $-4 $ and $3.$
$P(x)=-x(x+3)(x-4)$
Zeros: $-3, 0,4,$ all single.
y-intercept:$ (0,0)$
The leading factor of a third degree polynomial is negative.
The end behavior is as for $-x^{3}$ ($+\infty$ at the far left, $-\infty$ at the far right)..
Sketch:
On the far left, it falls from $+\infty,$
crosses x at $(-3,0)$, falls and turns
to cross x at $(0,0)$, rises and
turns to cross x again at $(0,2)$, falling
and continues falling to the far right.
