#### Answer

a. $\frac{p}{q}=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$
b. $x=-2$; see figure.
c. $x=-2, 1\pm\sqrt 7$

#### Work Step by Step

a. Given the function $f(x)=x^3-10x-12$, 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 $x=-2$ as a zero shown in the figure.
c. Based on the results from part-b, we have $f(x)=x^3-10x-12=(x+2)(x^2-2x-6)$
Thus the zeros are $x=-2, 1\pm\sqrt 7$