College Algebra, Chapter 4, 4.2, Section 4.2, Problem 28

Factor the polynomial $P(x) = x^3 + 2x^2 - 8x$ and use the factored form to find the zeros. Then sketch the graph.

$\begin{equation} \begin{aligned} P(x) &= x^3 + 2x^2 - 8x \\ \\ &= x (x^2 + 2x - 8) && \text{Factor out } x\\ \\ &= x(x+4)(x-2) && \text{Simplify} \end{aligned} \end{equation}$

Since the function has an odd degree of 3 and a positive 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
$x = 0, \quad x + 4 = 0$ and $x - 2 = 0$

Thus, the $x$-intercept are $x = 0, -4$ and 2