Answer
a, $\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$
b. $1$ is a zero
c. $x\displaystyle \in\{-3, 1, 4\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ 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, $a_{n}$. ------------------------ $f(x)=x^{3}-2x^{2}-11x+12$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$
b. Try for $x=1:$
$\begin{array}{lllll} \underline{1}| & 1 & -2 & -11 & 12\\
& & 1 & -1 & -12\\
& -- & -- & -- & --\\
& 1 & -1 & -12 & |\underline{0} \end{array}$
${1}$ is a zero,
$f(x)=(x-1)(x^{2} -x-12)$
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, $x^{2} -x-12$, $a=1, b=-1, c=-12$
$x=\frac{1 \pm \sqrt {(-1)^2-4\times 1 \times (-12)}}{2\times 1}$, $x=\frac{1 \pm \sqrt {49}}{2}$, $x=\frac{1 \pm 7}{2}$
thus,
$x=4$ or $x=-3$
The zeros of f satisfy $f(x)=0$
$x\displaystyle \in\{-3, 1, 4\}$
