Question

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

Short answer

The required interval is$(-3,0)\cup (1,\infty )$

Step-by-step solution

Step 1. Given Information 

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

This is given to us . we need to solve for x .

Step 2. Factorization

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

The zeros of the equation are 0, 1 & -3

Step 3.  Dividing the number line into intervals using the above zeros.

The intervals are :

$$\begin{gathered} (-\infty ,-3) \\ (-3,0) \\ (0,1) \\ (1,\infty ) \end{gathered}$$

Step 4. Evaluating the value of function 

$$\begin{gathered} 1.(-\infty ,-3) \\ letnumberbe-4 \\ valueoffunctionat-4=-20 \\ conditionnotsatisfied. \\ 2.(-3,0) \\ Letnumberbe-1 \\ valueoffunctionat-1=4 \\ Theconditionissatisfied. \\ 3.(0,1) \\ letnumberbe0.1 \\ valueoffunctionat0.1=-0.099 \\ conditionnotsatisfied. \\ 4.(0,\infty ) \\ Letnumberbe2 \\ valueoffunctionat2=10 \\ Theconditionissatisfied. \end{gathered}$$

Step 5. conclusion

The required interval is$(-3,0)\cup (1,\infty )$