(x+1)2(x-6)(x+4)≤0. inequation

Given the inequality:
$$2 \left(x + 1\right) \left(x - 6\right) \left(x + 4\right) \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \left(x + 1\right) \left(x - 6\right) \left(x + 4\right) = 0$$
Solve:
Given the equation:
$$2 \left(x + 1\right) \left(x - 6\right) \left(x + 4\right) = 0$$
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$x - 6 = 0$$
$$x + 4 = 0$$
$$2 x + 2 = 0$$
solve the resulting equation:
1.
$$x - 6 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = 6$$
We get the answer: x1 = 6
2.
$$x + 4 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = -4$$
We get the answer: x2 = -4
3.
$$2 x + 2 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$2 x = -2$$
Divide both parts of the equation by 2

x = -2 / (2)

We get the answer: x3 = -1
$$x_{1} = 6$$
$$x_{2} = -4$$
$$x_{3} = -1$$
$$x_{1} = 6$$
$$x_{2} = -4$$
$$x_{3} = -1$$
This roots
$$x_{2} = -4$$
$$x_{3} = -1$$
$$x_{1} = 6$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-4 + - \frac{1}{10}$$
=
$$- \frac{41}{10}$$
substitute to the expression
$$2 \left(x + 1\right) \left(x - 6\right) \left(x + 4\right) \leq 0$$
$$2 \left(- \frac{41}{10} + 1\right) \left(-6 + - \frac{41}{10}\right) \left(- \frac{41}{10} + 4\right) \leq 0$$

-3131      
------ <= 0
 500       

one of the solutions of our inequality is:
$$x \leq -4$$

 _____           _____          
      \         /     \    
-------•-------•-------•-------
       x2      x3      x1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq -4$$
$$x \geq -1 \wedge x \leq 6$$