Answer
a. $p:\qquad \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm12$
b. $-2$ is a root of the function
c. $x\in\{\ 1 \pm \sqrt 7, -2\}$
Work Step by Step
See The Rational Zero Theorem:
If $\displaystyle \frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients,
then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$f(x)=x^{3}-10x-12$
a. Candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm12$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm 3, \pm 4, \pm6, \pm12$
b. Try for $x=-2:$
$\begin{array}{lllll}
\underline{-2}| & 1 & 0 & -10 & -12\\
& & -2 & 4 & 12\\
& -- & -- & -- & --\\
& 1 & -2 & -6 & |\underline{0}
\end{array}$
${-2}$ is a zero,
$f(x)=(x+2)(x^{2} -2x-6)$
c. Using quadratic formula, $x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$.
we can solve for the quadratic $ax^2+bx+c$,
In our case of, $x^{2} -2x-6$, $a=1, b=-2, c=-6$
$x=\frac{2 \pm \sqrt {(-2)^2-4\times 1 \times (-6)}}{2\times 1}$,
$x=\frac{2 \pm 2\sqrt {7}}{2}$,
$x={1 \pm \sqrt {7}}$
The zeros of f satisfy $f(x)=0$
$x\displaystyle \in\{\ 1 \pm \sqrt 7, -2\}$
