College Algebra, Chapter 4, 4.5, Section 4.5, Problem 58

Determine all the zeros of the polynomial $P(x) = x^4 - x^2 + 2x + 2$.
To find the zeros of $P$, we set $x^4 - x^2 + 2x + 2 = 0$. The possible rational zeros of $P$ are the factors of 2 which are $\pm 1$ and $\pm 2$
Then by using synthetic division and trial and error,



Again, by using trial and error,



Thus,

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

If $(x+1)^2 = 0$, then $x = -1$ is a zero of $P$ with multiplicity 2. To determine the remaining zeros, we use quadratic formula

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

Hence, the factor of $P$ are $-1, 1 + i$ and $1 - i$