Answer
a. $\frac{p}{q}=\pm1,\pm2,\pm4,\pm\frac{1}{2}$
b. $x=\frac{1}{2}$; see figure.
c. $x=\frac{1}{2},1\pm\sqrt 5$
Work Step by Step
a. Given the function $f(x)=2x^3-5x^2-6x+4$, we have $p=\pm1,\pm2,\pm4$ and $q=\pm1,\pm2$; thus the possible rational zeros are $\frac{p}{q}=\pm1,\pm2,\pm4,\pm\frac{1}{2}$
b. Starting from the easiest number, test the possible rational zeros using synthetic division; we can find $x=\frac{1}{2}$ as a zero shown in the figure.
c. Based on the results from part-b, we have $f(x)=(x-\frac{1}{2})(2x^2-4x-8)=(2x-1)(x^2-2x-4)$
Thus the zeros are $x=\frac{1}{2},1\pm\sqrt 5$
