College Algebra, Chapter 4, 4.2, Section 4.2, Problem 32

Factor the polynomial $P(x) = x^5 - 4x^3$ and use the factored form to find the zeros. Then sketch the graph.

$\begin{equation} \begin{aligned} P(x) &= x^5 - 4x^3 \\ \\ &= x^3(x^2 - 9) && \text{Factor out } x^2\\ \\ &= x^3(x+3)(x-3) && \text{Difference of squares} \end{aligned} \end{equation}$

Since the function has an odd degree of 5 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 $y = 0$.
$0 = x^3(x+3)(x-3)$

By zero product property, we have
$x^3 = 0, \quad x + 3 = 0$ and $x - 3 = 0$

Thus, the $x$-intercept are $x =0, -3$ and 3