Answer
a. $\frac{p}{q}=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$
b. $x=1$ (see figure).
c. $x=-3,1,4$
Work Step by Step
a. Given the function
$f(x)=x^3-2x^2-11x+12$
we have
$p=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$
and
$q=\pm1$
Thus the possible rational zeros are
$\frac{p}{q}=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$
b. Starting from the easiest number, test the possible rational zeros using synthetic division. We can find $x=1$, as a zero shown in the figure.
c. Based on the results from part-b, we have
$f(x)=x^3-2x^2-11x+12=(x-1)(x^2-x+12)=(x-1)(x-4)(x+3)$
Thus the zeros are $x=-3,1,4$
