Answer
$\pm 1, \pm 2, \pm 3, \pm4, \pm6, \pm12, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{1}{6}, \pm\frac{2}{3}, \pm\frac{3}{2}, \pm\frac{4}{3}$
Work Step by Step
RECALL:
The possible rational zeros of a polynomial function is given by $\dfrac{p}{q}$ where:
$p$ = factor of the constant term
$q$ = factor of the leading coefficient
The given polynomial function has:
constant term = $12$
leading coefficient = $6$
The factors of the constant term are: $\pm1, \pm2, \pm 3, \pm4, \pm6, \pm12$
The factors of the leading coefficient are: $\pm 1, \pm2, \pm3, \pm6$
Thus, the possible rational zeros of the given polynomial function are:
$=\pm 1, \pm 2, \pm 3, \pm4, \pm6, \pm 12, \pm\frac{1}{2}, \pm \frac{1}{3}, \pm\frac{1}{6}, \pm\frac{2}{2}, \pm\frac{2}{3}, \pm\frac{2}{6}, \pm\frac{3}{2}, \frac{3}{3}, \pm\frac{3}{6}, \pm\frac{4}{2},\pm\frac{4}{3}, \pm\frac{4}{6}, \pm\frac{6}{2}, \pm\frac{6}{3}, \pm\frac{6}{6}, \pm\frac{12}{2}, \pm\frac{12}{2}, \pm\frac{12}{3}, \pm\frac{12}{6}$
Eliminate the duplicates to obtain:
$\\=\pm 1, \pm 2, \pm 3, \pm4, \pm6, \pm12, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{1}{6}, \pm\frac{2}{3}, \pm\frac{3}{2}, \pm\frac{4}{3},$
