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 $q=\pm1$
and the possible rational zeros are
$\frac{p}{q}=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$
b. Starting from the lowest, test the possible rational zeros using synthetic division. We can find $x=1$ as a zero as 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 $x=-3,1,4$
