College Algebra, Chapter 4, 4.4, Section 4.4, Problem 10

Find all possible rational zeros of $U(x) = 12x^5 + 6x^3 - 2x - 8$ using the rational zeros theorem (don't check to see which actually are zeros).

By the rational zeros theorem, the rational zeros of $U$ are of the form


$
\begin{equation}
\begin{aligned}

\text{possible rational zero of } U =& \frac{\text{factor of constant term}}{\text{factor of leading coefficient}}
\\
\\
=& \frac{\text{factor of 8}}{\text{factor of 12}}

\end{aligned}
\end{equation}
$


The factors of $8$ are $\pm 1, \pm 2, \pm 4, \pm 8$ and the factors of $12$ are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$. Thus, the possible rational zeros of $U$ are

$\displaystyle \pm \frac{1}{1}, \pm \frac{2}{1}, \pm \frac{4}{1}, \pm \frac{8}{1}, \pm \frac{1}{2}, \pm \frac{2}{2}, \pm \frac{4}{2}, \pm \frac{8}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}, \pm \frac{8}{3}, \pm \frac{1}{4}, \pm \frac{2}{4}, \pm \frac{4}{4}, \pm \frac{8}{4}, \pm \frac{1}{6}, \pm \frac{2}{6}, \pm \frac{4}{6}, \pm \frac{8}{6}, \pm \frac{1}{12}, \pm \frac{2}{12}, \pm \frac{4}{12}, \pm \frac{8}{12}$.

Simplifying the fractions and eliminating duplicates, we get

$\displaystyle \pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}$.