Answer
$\left\{-1,-\dfrac{1}{2},3\right\}$
Work Step by Step
We are given the equation:
$2x^3-3x^2-8x-3=0$
First we look for a rational solution in the set:
$\pm 1, \pm 3, \pm\dfrac{1}{2},\pm\dfrac{3}{2}$
Use synthetic division to look for a rational zero. We find $x=-1$ is a solution:
$(x+1)(2x^2-5x-3)=0$
Determine the zeros of the quadratic polynomial:
$x=\dfrac{5\pm\sqrt{(-5)^2-4(2)(-3)}}{2(2)}=\dfrac{5\pm 7}{4}$
$x_1=\dfrac{5-7}{4}=-\dfrac{1}{2}$
$x_2=\dfrac{5+7}{4}=3$
The solution set is:
$\left\{-1,-\dfrac{1}{2},3\right\}$
