Answer
$\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{1}{6},\pm\frac{2}{3},\pm\frac{5}{2},\pm\frac{5}{3},\pm\frac{5}{6},\pm\frac{10}{3},\pm1,\pm2,\pm5,\pm10$
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 $10$ and a leading coefficient of $-6$.
The possible factors $p$ of the constant term and $q$ of the leading coefficient are:
$p=\pm1,\pm2,\pm4,\pm5,\pm10$
$q=\pm1,\pm2,\pm3,\pm6$
Thus, the possible rational roots of $f(x)$ are:
$\frac{p}{q}=\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{1}{6},\pm\frac{2}{3},\pm\frac{5}{2},\pm\frac{5}{3},\pm\frac{5}{6},\pm\frac{10}{3},\pm1,\pm2,\pm5,\pm10$