Answer
$P(x)=-x(x+1)(2x-1)$
Zeros: $-1,\displaystyle \quad 0,\quad \frac{1}{2},$ all single.
Work Step by Step
Factor out $-x$
$P(x)=-x(2x^{2}+x-1)$
$p=\pm 1,\quad q=\pm 1,\pm 2$
$\displaystyle \frac{p}{q}$ are eligible for zeros of P(x). Synthetic division ...
\begin{array}{rrrr}
-1 \rceil & 2 & 1 & -1 \\
& & -2 & 1 \\
\hline & 2 & -1 & 0 \\
\end{array}
$x=-1 $ is a zero
$P(x)=-x(x+1)(2x-1)$
Zeros: $-1,\displaystyle \quad 0,\quad \frac{1}{2},$ 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 $(-1,0)$, falls and turns
to cross x at $(0,0)$, rises and
turns to cross x again at $(\displaystyle \frac{1}{2},0)$, falling
and continues falling to the far right.