Where did $-4x$ come from? I'm going over my quadratic equations for the ACT and I came across this quadratic:
$$(x – 2)^2  – 12$$
My teacher said we could have factored it out into this:
$$x^2 – 4x – 8$$
But I just don't understand where he got the $-4x$! Help?
 A: Same as others but with some colors
$(x-2)^2=\color{red}{(x-2)}\color{blue}{(x-2)}$
$\color{red}{(x-2)}\color{blue}{(x-2)}=\color{red}x\color{blue}{(x-2)}\color{red}{-2}\color{blue}{(x-2)}$
$\hspace{65 pt}=\underbrace{\color{red}x\times \color{blue}x}\hspace{5 pt}+\underbrace{\color{red}x \times \color{blue}{-2}}\hspace{5 pt}\underbrace{\color{red}{-2} \times \color{blue}x}\hspace{5 pt}\underbrace{ \color{red}{-2} \times \color{blue}{-2}}$
  $\hspace{70 pt}=x^2 \hspace{25 pt}-2x\hspace{15 pt}-2x\hspace{15 pt}+4$
$\hspace{65 pt}=x^2-4x+4$
A: $$(x-2)^2=(x-2)(x-2)=x(x-2)-2(x-2)=x^2-2x-2x+4=x^2-4x+4$$
A: $x^{2}-4x-8$=$(x-2)^{2}-4-8$=$(x-2)^{2}-12$
A: $$(x-2)^2 = (x - 2)(x-2) = x^2 - 2x -2x + (-2)(-2) = x^2 - 4x + 4$$
This is called expanding $(x-2)^2$. We factor $x^2 - 4x + 4$ when we write it as the product of its factors, in this case $(x-2)(x-2) = (x-2)^2$.
Now, 
$$(x-2)^2-12=(x^2-4x+4)-12=x^2-4x-8$$
You can find the zeros of the quadratic by setting $x^2 - 4x - 8 = 0 $, then using the quadratic formula, which will yield $x = 2\pm 2\sqrt 3$.
A: $$(x-2)^2=x^2-4x+4$$
$$(x-2)^2-12=x^2-4x+4-12=x^2-4x-8$$
