Answer
The inequality is valid for values between -1 and 0, and values more than 3 (not including them) i.e. $(-1,0)\cap (3,\infty)$
Work Step by Step
First, we are going to find the x-intercepts by equating to zero:
$x^3-2x^2-3x=0$
$x(x^2-2x-3)=0$
$x(x+1)(x-3)=0$
$x_1=-1$
$x_2=0$
$x_3=3$
These are the critical points. We are going to take four values: one less than -1; one between -1 and 0; one between 0 and 3; and one more than 3 to test in the original equation and check if the inequality is true or not:
First test with a value less than -1:
$(-2)^3-2(-2)^2-3(-2)\gt0$
$-8-8+6>0$
$-10>0 \rightarrow \text{ FALSE}$
Second test with a value between -1 and 0:
$(-0.5)^3-2(-0.5)^2-3(-0.5)\gt0$
$-0.125-0.5+1.5>0$
$0.875> \rightarrow \text{ TRUE}$
Third test with a value between 0 and 3:
$1^3-2(1)^2-3(1)\gt0$
$1-2-3>0$
$-4>0 \rightarrow \text{ FALSE}$
Fourth test with a value more than 3:
$4^3-2(4)^2-3(4)\gt0$
$64-32-12>0$
$20>0 \rightarrow \text{ TRUE}$
These tests show that the inequality $x^3-2x^2-3x\gt0$ is valid for values between -1 and 0, and values greater than 3 (not including them) i.e. $(-1,0)\cap (3,\infty)$
