Answer
$\left\{-5,-\dfrac{1}{3},2\right\}$
Work Step by Step
We are given the equation:
$9x^4+33x^3-71x^2-57x-10=0$
First try to find a rational solution in the set:
$\pm 1, \pm 2, \pm 5, \pm 10, \pm\dfrac{1}{3},\pm \dfrac{1}{9}, \pm\dfrac{2}{3},\pm \dfrac{2}{9}, \pm\dfrac{5}{3},\pm \dfrac{5}{9},\pm\dfrac{10}{3},\pm \dfrac{10}{9}$
Using synthetic division we get that $x_1=2$ is a solution:
$(x-2)(9x^3+51x^2+31x+5)=0$
We use synthetic division again for the cubic polynomial and we get that $x_2=-5$ is also a solution:
$(x-2)(x+5)(9x^2+6x+1)=0$
Factor the quadratic polynomial:
$(x-2)(x+5)(3x+1)^2=0$
Find all solutions:
$x_1=2$
$x_2=-5$
$3x+1=0\Rightarrow x_3=-\dfrac{1}{3},x_4=-\dfrac{1}{3}$
The solution set is:
$\left\{-5,-\dfrac{1}{3},2\right\}$
