College Algebra, Chapter 4, 4.5, Section 4.5, Problem 68

(a) Factor $P(x) = x^5 - 16x$ into linear and irreducible quadratic factors with real coefficients. (b) Factor $P$ completely into linear factors with complex coefficients.


$
\begin{equation}
\begin{aligned}
\text{a.) } P(x) &= x^5 - 16x\\
\\
&= x (x^4 - 16) && \text{Factor out } x\\
\\
&= x (x^2 - 4)( x^2 + 4) && \text{Difference of squares}
\end{aligned}
\end{equation}
$

The factor $(x^2 + 4)$ is irreducible, since it has no real zeros

b.) To get the complete factorization, we set $x^2 + 4 = 0$, so
$x = \pm 2i$
Thus,

$
\begin{equation}
\begin{aligned}
P (x) &= x^5 - 16x\\
\\
&= x(x^4 - 16)\\
\\
&= x (x^2 -4)(x^2 + 4)\\
\\
&= x(x^2 -4)(x-2i)(x+2i)
\end{aligned}
\end{equation}
$