Answer
$\frac{p}{q}= \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$
Work Step by Step
The rational root theorem (or rational root test, rational zero theorem, rational zero test or p/q theorem) states a constraint on rational solutions of a polynomial equation
${\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{0}=0}$ with integer coefficients $a_{i} \in Z$ and $a_{0}$, ${\displaystyle a_{0},a_{n}\neq 0}$.
The theorem states that each rational solution $x = p⁄q$, written in lowest terms so that $p$ and $q$ are relatively prime, satisfies:
$p$ is an integer factor of the constant term $a_{0}$, and
$q$ is an integer factor of the leading coefficient $a_{n}$.
In this case, $f(x)=2x^3+11x^2-7x-6$
$p= \pm 1, \pm 2, \pm 3, \pm 6$
$q= \pm 1, \pm 2$
$\frac{p}{q}= \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$
