College Algebra, Chapter 4, 4.5, Section 4.5, Problem 12

a.) Find all zeros of $P(x) = x^4 + 6x^2 + 9$ of $P$, real and complex

b.) Factor $P$ completely.



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


$
\begin{equation}
\begin{aligned}

P(x) =& x^4 + 6x^2 + 9
&& \text{Given}
\\
\\
=& w^2 + 6w + 9
&& \text{Let } w = x^2
\\
\\
=& (w + 3)^2
&& \text{Factor}
\\
\\
=& (x^2 + 3)^2
&& \text{Substitute } w = x^2

\end{aligned}
\end{equation}
$


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

Setting $(x^2 + 3)^2 = 0$, we get $x^2 + 3 = 0$ is a zero, so $x = \pm \sqrt{3} i$. Hence, the zeros of $P$ are $\sqrt{3} i$ and $- \sqrt{3} i$.

b.) By complete factorization,


$
\begin{equation}
\begin{aligned}

P(x) =& \left( x - \sqrt{3} i \right) \left[ x - \left( - \sqrt{3} i \right) \right]
\\
\\
=& \left( x - \sqrt{3} i \right) \left( x + \sqrt{3} i \right)

\end{aligned}
\end{equation}
$