How to Solve Inequalities of Degree 3 or More

First, you have to move all the terms over to the left-hand side. Then you need to factorize the cubic equation $P(x)$ that appears. You do this by guessing at a solution. When you are guessing at a solution, a natural starting point is the values $x=1,-1,2,-2,\dots$. By inserting these values into the cubic expression you find out that $P(1)=0$, which means $x=1$ is a zero of $P(x)$. Then you can use polynomial long division by dividing $P(x)$ by $x-1$, which will give you a quadratic expression. It looks like this:

$$\begin{array}{llll}\hfill x^3+6x^2-6& >2x^2-x& \hfill & \\ \hfill x^3+6x^2-6-2x^2+x& >0& \hfill & \\ \hfill x^3+4x^2+x-6& >0& \hfill & \end{array}$$

Solve this inequality like you would an equation. You do that by setting the left-hand side equal to zero. This is where you guess at a solution. Try $x=1$ first:

$$\begin{array}{llll}\hfill P(1)& ={(1)}^3+4{(1)}^2+(1)-6& \hfill & \\ \hfill & =1+4+1-6& \hfill & \\ \hfill & =0.& \hfill & \end{array}$$

Luckily, you only had to test one solution. You have now found $x=1$ to be a solution to $P(x)=0$, meaning that $x=1$ is a zero of $P(x)$. Then you can perform the polynomial long division $P(x)÷(x-1)$ to factorize $P(x)$ further:

Next, you factorize the solution to the polynomial long division. By inspection you can see that

$$x^2+5x+6=(x+2)(x+3).$$

That means the factorized left-hand side of the cubic inequality is

Make a sign chart for every factor in the expression and find out where $x^3+4x^2+x-6$ is positive and negative. Because the original inequality asked you to find where $x^3+6x^2-6>2x^2-x$, you’re looking for the intervals where the lines in the sign chart are solid.

The sign chart tells you that

$$x^3+6x^2-6>2x^2-x$$

when

$$x\in \left(-3,-2\right)\cup \left(1,\infty \right).$$