College Algebra, Chapter 4, 4.5, Section 4.5, Problem 6

a.) Find all zeros of $P(x) = x^5 + 9x^3 $ of $P$, real and complex

b.) Factor $P$ completely.



a.) We first factor $P$ as follows.


$
\begin{equation}
\begin{aligned}

P(x) =& x^5 + 9x^3
&& \text{Given}
\\
\\
=& x^3 (x^2 + 9)
&& \text{Factor out } x^3

\end{aligned}
\end{equation}
$


We find the zeros of $P$ by setting each factor equal to :

Setting $x^3 = 0$, we see that $x = 0$ is a zero. More over, setting $x^2 + 9 = 0$, we get $x^2 = -9$, so $x = \pm 3i$. So the zeros of $P$ are $0, 3i$ and $-3i$.

b.) Since the zeros are $0, 3i$ and $-3i$, by the complex Factorization Theorem $P$ factors as


$
\begin{equation}
\begin{aligned}

P(x) =& x (x - 3i) [x - (-3i)]
\\
\\
=& x (x - 3i)(x + 3i)

\end{aligned}
\end{equation}
$