College Algebra, Chapter 4, 4.2, Section 4.2, Problem 40

Factor the polynomial $P(x) = x^6 - 2x^3 + 1$ 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^6 - 2x^3 + 1\\
\\
0 &= w^2 - 2w + 1 && \text{Let } w = x^3\\
\\
0 &= (w-1)^2 && \text{Perfect Square} \\
\\
w &= 1 && \text{Substitute} w = x^3\\
\\
x^3 &= 1
\end{aligned}
\end{equation}
$



Thus, the $x$-intercept are $x = 1$