Question

Solve the inequality algebraically.

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

Short answer

The required interval set is :

$(-3,0)\cup (1,\infty )$

Step-by-step solution

Step 1. Given inequality 

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

we need to solve for x.

Step 2. Factorization

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

The zeros of the above equation are : 0,1,-3

Step 3. Dividing the number line into 4 intervals 

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

These are the intervals

Step 4. Evaluating the value of function at different intervals

$$\begin{gathered} 1.(-\infty ,-3) \\ Letustakenumber-4 \\ Thevalueofthefunctionat-4=-20 \\ theconditionisnotsatisfied. \\ 2.(-3,0) \\ Letustakenumber-1 \\ Thevalueofthefunctionat-1=4 \\ theconditionissatisfied. \\ 3.(0,1) \\ Letustakenumber0.1 \\ Thevalueofthefunctionat0.1=-0.279 \\ theconditionisnotsatisfied. \\ 4.(1,\infty ) \\ Letustakenumber2 \\ Thevalueofthefunctionat2=10 \\ theconditionissatisfied. \end{gathered}$$

Step 5. Conclusion

The required interval set is :

$(-3,0)\cup (1,\infty )$