College Algebra, Chapter 4, 4.4, Section 4.4, Problem 80

Determine all rational zeros of the polynomial $P(x) = 6x^4 - 7x^3 - 8x^2 + 5x$ 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.

$\begin{equation} \begin{aligned} P(x) &= 6x^4 - 7x^3 - 8x^2 + 5x\\ \\ P(x) &= x\left( 6x^3 - 7x^2 - 8x + 5 \right) && \text{Factor out } x \end{aligned} \end{equation}$

This means that is a root of $P$. We now factor $6x^3 - 7x^2 - 8x + 5$. The possible rational zeros are $\displaystyle \pm \frac{1}{6}, \pm \frac{1}{3}, \pm \frac{1}{2}, \pm \frac{5}{6}, \pm 1, \pm \frac{5}{3}, \pm \frac{5}{2}, \pm 5$. We check the positive candidates first, beginning with the smallest
Using Synthetic Division



So, $\displaystyle \frac{1}{2}$ is a zero and $\displaystyle P(x) = x \left( x - \frac{1}{2} \right) \left( 6x^2 - 4x - 10 \right)$. We now factor the quotient $6x^2 - 4x - 10$ by using trial and error
Therefore,

$\begin{equation} \begin{aligned} P(x) &= x \left( x - \frac{1}{2} \right) (2x + 2) (3x - 5) && \text{Factor out } 2\\ \\ P(x) &= 2x \left( x - \frac{1}{2} \right) (x + 1) (3x - 5) \end{aligned} \end{equation}$

This means that the zeros of $P$ are, $\displaystyle -1, 0, \frac{1}{2} \text{ and } \frac{5}{3}$.