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?

Use the remainder theorem for polynomials. 
$$
f(a) = 0 \implies f(x) = (x-a)p(x)
$$
Where $p(x)$ is reduced a degree. To find the $2$ you would trail solutions to find a root using a change of sign. 
$\textbf{update}$
Since we have found one root we can assume that the we have an equation of the form 
$$
(x+2)\left(ax^2+bx+c\right)
$$
Thus we have to compare the coefficents leading to
$$
a = 1\\
2a+ b = 4\\
2b + c = 1\\
2c = -6
$$
