#### Answer

The possible rational zeroes of the given function are $x=\left\{ \pm \frac{1}{2},\pm \frac{3}{2},\pm 1,\pm 2,\pm 3,\pm 6 \right\}$.

#### Work Step by Step

The given polynomial is $f\left( x \right)=2{{x}^{3}}+11{{x}^{2}}-7x-6$.
Determine the values of p and q, where p are the factors of the constant in the polynomial and q are the factors of the leading coefficient of the polynomial.
Thus:
$p=\pm 1,\pm 2,\pm 3,\pm 6$
$q=\pm 1,\pm 2$
Calculate $\frac{p}{q}$ and obtain the roots of the given function as:
$\frac{p}{q}=\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2}$.
Thus, the possible rational zeroes of the function are $x=\left\{ \pm \frac{1}{2},\pm \frac{3}{2},\pm 1,\pm 2,\pm 3,\pm 6 \right\}$.