College Algebra, Chapter 4, 4.5, Section 4.5, Problem 60

Determine all the zeros of the polynomial $P(x) = 4x^4 + 2x^3 - 2x^2 - 3x - 1$.
To determine the zeros of $P$, we set $4x^4 + 2x^3 - 2x^2 - 3x - 1 = 0$. Based from the theorem, the possible rational zeros of $P$ are the factors of 1 divided by the factors of the leading coefficient 4.
We have $\displaystyle \pm \frac{1}{1}, \pm \frac{1}{2}$ and $\displaystyle \pm \frac{1}{4}$. Then, by using synthetic division and trial and error




Similarly, by applying synthetic division to the possible rational roots
$\displaystyle \pm \frac{1}{1}, \pm \frac{2}{1}, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{2}{2}, \pm \frac{2}{4}$, by simplifying, we have $\displaystyle \pm 1, \pm 2, \pm \frac{1}{2} \text{ and } \pm \frac{1}{4}$. Then by trial and error



Then,

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

To find the remaining zeros of $P$, we use quadratic formula

$
\begin{equation}
\begin{aligned}
x &= \frac{-(4) \pm \sqrt{4^2 - 4(4)(2)}}{2(4)} \\
\\
&= \frac{-4 \pm \sqrt{-16}}{8}\\
\\
&= \frac{-1 \pm i }{2}
\end{aligned}
\end{equation}
$

Thus, the zeros of $P$ are $\displaystyle \frac{-1}{2}, 1, \frac{-1+i}{2} \text{ and } \frac{-1-i}{2}$