#### Answer

$\pm \dfrac{1}{2}, \pm \dfrac{1}{4}, \pm \dfrac{3}{4}, \pm \dfrac{3}{2}, \pm 1, \pm 2, \pm 3, \pm 6$

#### 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: $\dfrac{m}{n}$.
We see from the given polynomial function that it has a constant term of $6$ and a leading coefficient of $-4$.
The possible factors $m$ of the constant term and $n$ of the leading coefficient are: $m=\pm 1, \pm 2, \pm 3, \pm 6$ and $n=\pm 1, \pm 2, \pm 4$,
Therefore, the possible rational roots of $f(x)$ are:
$\dfrac{m}{n}=\pm \dfrac{1}{2}, \pm \dfrac{1}{4}, \pm \dfrac{3}{4}, \pm \dfrac{3}{2}, \pm 1, \pm 2, \pm 3, \pm 6$