How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$? How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$?
I get that you can take out the $x$ like so: $x^3+4x^2+x-6=x(x^2+4x+1)-6$ but how do you get the $2$ from here?

Knowing that $(x+2)$ is a factor, you know that the solution will be of the form
$$x^3+4x^2+x-6=(x+2)(ax^2+bx+c).$$
Execute the product 
$$(x+2)(ax^2+bx+c)=ax^3+(2a+b)x^2+(2b+c)x+(2c).$$
Identify the coefficients
$$a=1,2a+b=4,2b+c=1,2c=-6.$$
Solve and get
$$a=1,b=2,c=-3,$$
so that
$$x^3+4x^2+x-6=(x+2)(x^2+2x-3).$$
