College Algebra, Chapter 4, 4.5, Section 4.5, Problem 56

Determine all the zeros of the polynomial $P(x) = x^5 + x^3 + 8x^2 + 8 $.
To find the zeros, we set $x^5 + x^3 + 8x^2 + 8 = 0$. Then,

$
\begin{equation}
\begin{aligned}
(x^5 + x^3) + (8x^2 + 8) &= 0 && \text{Group terms}\\
\\
x^3(x^2+1)+8(x^2+1) &= 0 && \text{Factor out $x^3$ and $8$}\\
\\
(x^2+1)(x^3+8) &= 0 && \text{Factor out } x^2 + 1
\end{aligned}
\end{equation}
$

If $(x^2 + 1) = 0$, then $x = \pm i$. To solve the remaining zeros of $P$, we use synthetic division to simplify $(x^3+8) = 0$. So by trial and error,
Again, by using synthetic division



Thus,

$
\begin{equation}
\begin{aligned}
P(x) &= x^5 + x^3 + 8x^2 + 8 \\
\\
&= (x^2+1)(x^3+8)\\
\\
&= (x^2+1)(x+2)(x^2-2x+4)
\end{aligned}
\end{equation}
$


Then, by using quadratic formula...

$
\begin{equation}
\begin{aligned}
x &= \frac{-(-2)\pm \sqrt{(-2)^2 - 4(1)(4)}}{2(1)}\\
\\
&= \frac{2 \pm \sqrt{-12}}{2} = 1 \pm \sqrt{3}i
\end{aligned}
\end{equation}
$

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