College Algebra, Chapter 4, 4.2, Section 4.2, Problem 34

Factor the polynomial $P(x) = x^3 + 3x^2 - 4x - 12$ and use the factored form to find the zeros. Then sketch the graph.
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 $y = 0$.
$0 = x^3 + 3x^2 - 4x - 12$


$\begin{equation} \begin{aligned} 0 &= \left( x^3 + 3x^2 \right) - (4x + 12) && \text{Group terms}\\ \\ 0 &= x^2(x+3) - 4(x+3) && \text{Factor out $x^2$ and 4}\\ \\ 0 &= (x^2-4)(x+3) && \text{Factor out } x +3 \end{aligned} \end{equation}$



By zero product property, we have
$x^2 - 4 = 0$ and $x + 3 = 0$

Thus, the $x$-intercept are $x = -2, 2$ and $-3$