91Ó°ÊÓ

we have a given inequality

$x^3-2x^2-3x>0$

Zeros of inequality

$$\begin{gathered} f\left(x\right)=x^3-2x^2-3x>0are \\ f\left(x\right)=0 \\ {x}^3-2x^2-3x=0 \\ x(x^2-2x-3)=0 \\ x(x-3)(x+1)=0 \\ x=0 \\ x=3 \\ x=-1 \end{gathered}$$

Now we use the zeros to separate the real number line into intervals.

$(-∞,-1)(-1,0)(0,3)(3,∞)$

Now we select a test number in each interval found in Step 3 and evaluate at each number to determine if $f\left(x\right)=x^3-2x^2-3x=0$is positive or negative.

In the interval $(-∞,-1)$we chose -2 where f isnegative

In the interval $(-1,0)$we chose -0.5 , where f is positive.

In the interval (0,3) we chose 2.5 , where f is negative.

In the interval $(3,∞)$we chose 4 , where f is positive.

We know that our required inequality is $f\left(x\right)>0$

Here the inequality is not strict $(â‰\yen or≤)$ so we have to exclude the solutions of $f\left(x\right)=0$in the solution set.

So we want the interval where f is positive.

So the required solution set is$(-1,0)∪(3,∞)$