## Precalculus (10th Edition)

$\left\{-1,-\dfrac{1}{2},3\right\}$
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: $\left\{-1,-\dfrac{1}{2},3\right\}$