Answer
a. $\frac{p}{q}=\pm1,\pm\frac{1}{2}$
b. $x=\frac{1}{2}$, see figure.
c. $x=\frac{1}{2},\frac{-1\pm\sqrt 5}{2}$
Work Step by Step
a. Given the function $f(x)=2x^3+x^2-3x+1$, we have $p=\pm1$ and $q=\pm1,\pm2$
Thus the possible rational zeros are
$\frac{p}{q}=\pm1,\pm\frac{1}{2}$
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
$f(x)=2x^3+x^2-3x+1=(x-\frac{1}{2})(2x^2+2x-2)=(2x-1)(x^2+x-1)$
Thus the zeros are $x=\frac{1}{2},\frac{-1\pm\sqrt 5}{2}$
