a.) Find all zeros of $P(x) = x^4 - x^2 - 2$ of $P$, real and complex
b.) Factor $P$ completely.
a.) We first factor $P$ as follows.
$
\begin{equation}
\begin{aligned}
P(x) =& x^4 - x^2 - 2
&& \text{Given}
\\
\\
=& (w - 2)(w + 1)
&& \text{Factor}
\\
\\
=& (x^2 - 2)(x^2 + 1)
&& \text{Substitute } w = x^2
\end{aligned}
\end{equation}
$
We find the zeros of $P$ by setting each factor equal to :
Setting $x^2 - 2 = 0$, we get $x^2 = 2$ is a zero. More over, setting $x^2 + 1 = 0$, we get $x^2 = -1$, so $x = \pm i$ is a zero. Hence, the zeros of $P$ are $\sqrt{2}, - \sqrt{2}, i$ and $-I$.
b.) By complete factorization,
$
\begin{equation}
\begin{aligned}
P(x) =& (x - \sqrt{2}) \left[ x - \left( - \sqrt{2} \right) \right] [x - i] \left[ x - (-i) \right]
\\
\\
=& \left(x - \sqrt{2} \right) \left( x + \sqrt{2} \right) (x - i) (x + i)
\end{aligned}
\end{equation}
$
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