# How do you solve x^2(4-x)(x+6)<0?

The inequality is TRUE for values of x:
$x < - 6 \text{ }$ OR $\text{ } x > 4$

#### Explanation:

Since by solving for the values of x for each factor, we are going to have values $x = - 6$ and $x = 0$ and $x = 4$

The intervals are $\left(- \infty , - 6\right)$ and $\left(- 6 , 0\right)$ and $\left(0 , 4\right)$ and $\left(4 , + \infty\right)$

Let us use test points for each interval

For $\left(- \infty , - 6\right)$ , let us use $- 7$

For $\left(- 6 , 0\right)$ , let us use $- 2$

For $\left(0 , 4\right)$ , let us use $+ 1$

For $\left(4 , + \infty\right)$ , let us use $+ 5$

Let us do each test

At $x = - 7 \text{ }$the value$\text{ " " "x^2(4-x)(x+6)<0" }$TRUE
At $x = - 2 \text{ }$the value$\text{ " " "x^2(4-x)(x+6)<0" }$FALSE
At $x = + 1 \text{ }$the value$\text{ " " "x^2(4-x)(x+6)<0" }$FALSE
At $x = + 5 \text{ }$the value$\text{ " " "x^2(4-x)(x+6)<0" }$TRUE

Conclusion:

The inequality is TRUE for the following intervals
$\left(- \infty , - 6\right)$ and $\left(4 , + \infty\right)$

OR

The inequality is TRUE for values of x:
$x < - 6$ OR $x > 4$

God bless....I hope the explanation is useful.