I'm stuck trying to factor $x^2-4$ to $(x-2)(x+2)$

Question

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.

Answer 1

\begin{align} x^2-4&=x^2-2x+2x-4\\ &=x(x-2)+2(x-2)\\ &=(x+2)(x-2) \end{align}

Answer 2

$$x^2-4=x^2+2x-2x-4$$

$$=(x^2+2x) - (2x+4)$$

$$=x(x+2) - 2(x+2)$$

$$(x+2)(x-2).$$

Answer 3

Alternatively, recall the general formula $$a^2-b^2=(a+b)(a-b).$$ This can be seen by expanding the RHS, or by writing $$a^2-b^2=a^2-ab+ab-b^2=a(a-b)+b(a-b)=(a+b)(a-b).$$ Then, with $a=x$ and $b=2$, we have $$x^2-4=x^2-2^2=(x+2)(x-2).$$

Answer 4

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

$\tag {ANS} x^2 - 4 = (x+2) (x-2)$