College Algebra, Chapter 4, 4.2, Section 4.2, Problem 38

Factor the polynomial $P(x) = x^4 - 2x^3 + 8x - 16$ and use the factored form to find the zeros. Then sketch the graph.
Since the function has an even degree of 4 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 $y = 0$.


$\begin{equation} \begin{aligned} 0 &= x^4 - 2x^3 + 8x - 16\\ \\ 0 &= x^4 - 2x^3 + (8x - 16) && \text{Group terms}\\ \\ 0 &= x^3(x-2)+8(x-2) && \text{Factor out } x -2\\ \\ 0 &= (x^3+8)(x-2) && \text{Factor out } x^3 + 8 \end{aligned} \end{equation}$



By zero product property, we have
$x^3 + 8 = 0$ and $x - 2 = 0$

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