College Algebra, Chapter 4, 4.2, Section 4.2, Problem 30

Factor the polynomial $P(x) = -2x^3 - x^2 + x$ and use the factored form to find the zeros. Then sketch the graph.

$
\begin{equation}
\begin{aligned}
P(x) &= -2x^3 - x^2 + x \\
\\
&= -x (-2x^2 + x - 1) && \text{Factor out } -x\\
\\
&= -x(2x-1)(x+1) && \text{Simplify}
\end{aligned}
\end{equation}
$

Since the function has an odd degree of 3 and a negative leading coefficient, its end behaviour is $y \rightarrow \infty \text{ as } x \rightarrow -\infty \text{ and } y \rightarrow -\infty \text{ as } x \rightarrow \infty$. To find the $x$ intercepts (or zeros), we set $y = 0$.
$0 = -x(x+4)(x-2)$

By zero product property, we have
$-x = 0, \quad 2x -1 = 0$ and $x + 1 = 0$

Thus, the $x$-intercept are $\displaystyle x = 0, \frac{1}{2}$ and 2