College Algebra, Chapter 4, 4.5, Section 4.5, Problem 14

a.) Find all zeros of $P(x) = x^3 - 8$ of $P$, real and complex

b.) Factor $P$ completely.



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

The possible rational zeros are the factors of $8$, which are $\pm 1, \pm 2, \pm 4$ and $\pm 8$. Using Synthetic Division, by trial and error.







We find that $2$ is a zero, and the polynomial factors as.

$P(x) = (x - 2)(x^2 + 2x + 4)$

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

Setting $x - 2= 0$, we see that $x = 2$ is a zero. More over, setting $x^2 + 2x + 4 = 0$, by using Quadratic Formula


$
\begin{equation}
\begin{aligned}

x =& \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\\
\\
=& \frac{-2 \pm \sqrt{2^2 - 4(1)(4)}}{2(1)}
\\
\\
=& \frac{-2 \pm \sqrt{-12}}{2}
\\
\\
=& \frac{-2 \pm \sqrt{12} i}{2}
\\
\\
=& \frac{-2 \pm 2 \sqrt{3} i}{2}
\\
\\
=& -1 \pm \sqrt{3} i


\end{aligned}
\end{equation}
$


Hence, the zeros of $P$ are $2, -1 + \sqrt{3} i$ and $-1 - \sqrt{3} i$.

b.) By complete factorization,


$
\begin{equation}
\begin{aligned}

P(x) =& (x - 2) \left[ x - \left( -1 + \sqrt{3} i \right) \right] \left[ x - \left( - 1 - \sqrt{3} i \right) \right]
\\
\\
=& (x - 2) \left[ x + \left( 1 - \sqrt{3} i \right) \right] \left[ x + \left( 1 + \sqrt{3} i \right) \right]

\end{aligned}
\end{equation}
$