College Algebra, Chapter 4, 4.4, Section 4.4, Problem 82

Determine all rational zeros of the polynomial $P(x) = 8x^5 - 14x^4 - 22x^3 + 57x^2 - 35x + 6$ and then find the irrational zeros, if any. Whenever appropriate, use the Rational Zeros Theorem, the Upper and Lower Bounds Theorem, Descartes' Rule of Signs, the quadratic formula or other factoring techniques.
The possible rational zeros of $P$ are $\displaystyle \pm \frac{1}{4}, \pm \frac{3}{8}, \pm \frac{1}{2}, \pm \frac{3}{4}, \pm 1, \pm \frac{3}{2}, \pm 2, \pm 3, \pm 6$. We check the positive candidates first, beginning with the smallest
Using Synthetic Division, we have



So, $\displaystyle \frac{3}{4}$ is a zero and $\displaystyle P(x) = \left( x - \frac{3}{4} \right) \left( 8x^4 - 8x^3 - 28x^2 + 36x - 8 \right)$ or $\displaystyle P(x) = 4\left( x - \frac{3}{4} \right)\left( 2x^4 - 2x^3 - 7x^2 + 9x - 2 \right)$. We now factor the quotient $2x^4 - 2x^3 - 7x^2 + 9x - 2$. The possible rational zeros of $P$ are $\displaystyle \pm \frac{1}{2}, \pm 1, \pm 2$. Using Synthetic Division, we get



So, $1$ is a zero and $\displaystyle P(x) = 4\left( x - \frac{3}{4} \right) (x-1) \left( 2x^3 - 7x + 2 \right)$. We now factor the quotient $ 2x^3 - 7x + 2$. We still have same list of possible rational zeros except $\displaystyle \frac{1}{2}$ is elminated. Using Synthetic Division, we get


So, $-2$ is a zero and $\displaystyle P(x) = 4\left( x - \frac{3}{4} \right) (x-1)(x+2) \left( 2x^2 + 4x -1 \right)$. We now factor the quotient $ 2x^2 + 4x -1 $. By using quadratic formula,


$
\begin{equation}
\begin{aligned}
x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\
\\
x &= \frac{-4 \pm \sqrt{(4)^2 - 4(2)(-1)}}{2(2)}\\
\\
x &= \frac{-2\pm\sqrt{6}}{2}
\end{aligned}
\end{equation}
$

Therefore, $\displaystyle P(x) = 4\left( x - \frac{3}{4} \right) (x - 1)(x + 2) \left( x+\frac{2+ \sqrt{6}}{2} \right) \left( x + \frac{2-\sqrt{6}}{2} \right)$
This means that the zeros of $P$ are, $\displaystyle \frac{3}{4}, 1, -2, \frac{-2+\sqrt{6}}{2} \text{ and } \frac{-2-\sqrt{6}}{2}$.