I'm stuck trying to factor $x^2-4$ to $(x-2)(x+2)$ I am trying to understand each step in order to get from  $x^2-4$ to $(x-2)(x+2)$
I start from here and got this far...
$x^2-4 =$
$x*x-4 =$
$x*x+x-x-4 =$
$x*x+x-2+2-x-4 =$
$x*x+x-2+2-(x+4) =$ 
After this I try
$x(x-2)+2-(x+4) =$
and this clearly does not even equal the other factorings. I thought the $x$ could be factored out. I'm confused.
I know I can just insert $a^2-b^2$ into the difference of squares formula like so $(a-b)(a+b)$ but I am practicing factoring. I'm just curious to see each and every step of the factoring.

If $x^2 - 4$ can be factored then there exist $a, b \in \Bbb R$ such that
$\quad x^2 - 4 = (x-a) (x-b)$
Now multiplying out the rhs and collecting terms,
$\quad (x-a) (x-b) = x^2 - (a+b)x + ab$
So we need to solve the system of equations
$\tag 1 a + b = 0$
$\tag 2 ab = -4$
From equation $\text{(1)}$ we have
$\quad a = -b$
and plugging that into equation $\text{(2)}$,
$\quad -b^2 = -4$
We have two solutions for $b$.
If $b = 2$ then $a = -2$ and so
$\tag {ANS} x^2 - 4 = (x+2) (x-2)$
