## Precalculus (10th Edition)

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