Answer
$\\-1, 1, -2, 2, -4, 4, -8, 8, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{3}, \frac{1}{3}, -\frac{1}{4}, \frac{1}{4}, -\frac{1}{6}, \frac{1}{6}, -\frac{1}{12}, \frac{1}{12}, -\frac{2}{3}, \frac{2}{3}, -\frac{4}{3}, \frac{4}{3}, -\frac{8}{3}, \frac{8}{3}$
Work Step by Step
Note that:
The factors of $-8$ are $\pm1, \pm2, \pm4, \pm8$
The factors of $12$ are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$
Thus,
The possible values of p are $\pm1, \pm2, \pm4, \pm8$
The possible values of q are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$
Therefore the possible rational zeros $\frac{p}{q}$ of R(x) are:
$\\-1, 1, -2, 2, -4, 4, -8, 8, -\frac{1}{2}, \frac{1}{2}, -\frac{1}{3}, \frac{1}{3}, -\frac{1}{4}, \frac{1}{4}, -\frac{1}{6}, \frac{1}{6}, -\frac{1}{12}, \frac{1}{12}, -\frac{2}{3}, \frac{2}{3}, -\frac{4}{3}, \frac{4}{3}, -\frac{8}{3}, \frac{8}{3}$
