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 $1-\sqrt{3}i$