College Algebra, Chapter 4, 4.4, Section 4.4, Problem 8

Find all possible rational zeros of $S(x) = 6x^4 - x^2 + 2x + 12$ using the rational zeros theorem (don't check to see which actually are zeros).

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


$
\begin{equation}
\begin{aligned}

\text{possible rational zero of } S =& \frac{\text{factor of constant term}}{\text{factor of leading coefficient}}
\\
\\
=& \frac{\text{factor of 12}}{\text{factor of 6}}

\end{aligned}
\end{equation}
$


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

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

Simplifying the fractions and eliminating duplicates, we get

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