Answer
a. $\frac{p}{q}=\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2}$
b. $x=\frac{1}{2}$ see figure.
c. $x=-2,\frac{1}{2},3$
Work Step by Step
a. Given the function
$f(x)=2x^3-3x^2-11x+6$
we have
$p=\pm1,\pm2,\pm3,\pm6$ and $q=\pm1,\pm2$
and the possible rational zeros are
$\frac{p}{q}=\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2}$
b. Starting from the lowest, test the possible rational zeros using synthetic division. We can find $x=\frac{1}{2}$ as a zero as shown in the figure.
c. Based on the results from part-b, we have
$f(x)=2x^3-3x^2-11x+6=(x-\frac{1}{2})(2x^2-2x-12)=(2x-1)(x^2-x-6)=(2x-1)(x-3)(x+2)$
Thus the zeros are $x=-2,\frac{1}{2},3$
