Answer
$-1, 1, -2, 2, -3, 3, -4, 4, -6, 6, -12, 12, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{3}, \frac{1}{3}, -\frac{1}{6}, \frac{1}{6}, -\frac{2}{3}, \frac{2}{3}, -\frac{3}{2}. \frac{3}{2}, -\frac{4}{3}, \frac{4}{3}$
Work Step by Step
Note that:
The factors of $12$ are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$.
The factors of $6$ are $\pm1, \pm2, \pm3, \pm6$
Thus,
The possible values of $p$ are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$.
The possible values of $q$ are $\pm1, \pm2, \pm3, \pm6$
Therefore the possible rational zeros $\frac{p}{q}$ of $R(x)$ are:
$\\-1, 1, -2, 2, -3, 3, -4, 4, -6, 6, -12, 12, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{3}, \frac{1}{3}, -\frac{2}{3}, \frac{2}{3}, -\frac{3}{2}. \frac{3}{2}, -\frac{4}{3}, \frac{4}{3}-\frac{1}{6}, \frac{1}{6}$