Answer
$\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{9}{2},\pm\frac{1}{6},\pm\frac{3}{2},\pm1,\pm3,\pm9$
Work Step by Step
If $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient, then the potential zeros can be gained by the possible combinations in $\frac{p}{q}$. The given polynomial function has a constant term of $9$ and a leading coefficient of $6$.
The possible factors $p$ of the constant term and $q$ of the leading coefficient are: $p=\pm1,\pm3,\pm9$
$q=\pm1,\pm2,\pm3,\pm6$
hus, the possible rational roots of $f(x)$ are: $\frac{p}{q}=\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{9}{2},\pm\frac{1}{6},\pm\frac{3}{2},\pm1,\pm3,\pm9$