Answer
The possible rational roots of $f(x)$ are:
$\pm 1, \pm \dfrac{1}{2}, \pm \dfrac{1}{4},\pm \dfrac{1}{3}, \pm \dfrac{1}{6}, \pm \dfrac{1}{12}, \pm 3, \pm \dfrac{3}{2}, \pm \dfrac{3}{4}$
Work Step by Step
Let us consider that $m$ is a factor of the constant term and $n$ is a factor of the leading coefficient. Then the potential zeros can be expressed by the possible combinations as: $\dfrac{m}{n}$.
The possible factors $m$ of the constant term and $n$ of the leading coefficient are: $m=\pm 1, \pm 3$ and $n=\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$,
Therefore, the possible rational roots of $f(x)$ are:
$\dfrac{m}{n}=\pm 1, \pm \dfrac{1}{2}, \pm \dfrac{1}{4},\pm \dfrac{1}{3}, \pm \dfrac{1}{6}, \pm \dfrac{1}{12}, \pm 3, \pm \dfrac{3}{2}, \pm \dfrac{3}{4}$
