Answer
a. $\frac{p}{q}=\pm1,\pm2,\pm\frac{1}{2}$
b. $x=-2$; see figure.
c. $x=-2,\frac{-1\pm i}{2}$
Work Step by Step
a. Given the function $f(x)=2x^3+6x^2+5x+2$, 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 $x=-2$ as a zero (shown in the figure).
c. Based on the results from part-b, we have
$f(x)=2x^3+6x^2+5x+2=(x+2)(2x^2+2x+1)$
Thus the zeros are $x=-2,\frac{-1\pm i}{2}$
