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