Answer
a. $\frac{p}{q}=\pm1,\pm2,\pm3,\pm6$
b. $x=-1$; see figure.
c. $x=-1,\frac{-3\pm\sqrt {33}}{2}$
Work Step by Step
a. Given the function $f(x)=x^3+4x^2-3x-6$, we have $p=\pm1,\pm2,\pm3,\pm6$ and $q=\pm1$; thus the possible rational zeros are $\frac{p}{q}=\pm1,\pm2,\pm3,\pm6$
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 $f(x)=x^3+4x^2-3x-6=(x+1)(x^2+3x-6)$
Thus the zeros are $x=-1,\frac{-3\pm\sqrt {33}}{2}$
