Answer
$a_{0},\ \quad a_{n}$
$\displaystyle \pm 1,\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{1}{6},\pm 2,\displaystyle \pm\frac{2}{3},\pm 5,\pm\frac{5}{2},\pm\frac{5}{3},\pm\frac{5}{6},\pm 10,\pm\frac{10}{3}$
Work Step by Step
This is the Rational Zeros Theorem:
... every rational zero of $P$ is of the form $\displaystyle \frac{p}{q}$
where $p$ and $q$ are integers and
$p$ is a factor of the constant coefficient $a_{0}$
$q$ is a factor of the leading coefficient $a_{n}$
In $P(x)=6x^{3}+5x^{2}-19x-10,\quad a_{0}=-10,\quad a_{n}=6$
candidates for p: $\pm 1,\pm 2,\pm 5,\pm 10$
candidates for q: $\pm 1,\pm 2,\pm 3,\pm 6$
Possible rational roots:
$\displaystyle \pm 1,\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{1}{6},\pm 2,\displaystyle \pm\frac{2}{3},\pm 5,\pm\frac{5}{2},\pm\frac{5}{3},\pm\frac{5}{6},\pm 10,\pm\frac{10}{3}$
