#### Answer

a. $\frac{p}{q}=\pm1,\pm2,\pm4$
b. $x=-1$ (see figure).
c. $x=-1,4$

#### Work Step by Step

a. Given the function $f(x)=x^3-2x^2-7x-4$, we have
$p=\pm1,\pm2,\pm4$ and $q=\pm1$
Thus the possible rational zeros are
$\frac{p}{q}=\pm1,\pm2,\pm4$
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
$f(x)=x^3-2x^2-7x-4=(x+1)(x^2-3x-4)=(x+1)(x-4)(x+1)$
Thus the zeros are $x=-1,4$