<s>[INST] What are the steps to factor $x^2 - 1$ into $(x+1)(x-1)$? Does $(x+1)(x-1) = x^2+1x-1x-1$? If so where are the $+1x$ and the $-1x$ when it is being factored from $x^2-1$ into $(x+1)(x-1)$?
What exactly are we dividing $x^2-1$ by to get $(x+1)(x-1)$ and how did you know what to divide it by?
 [/INST] Method 0:
$+1x - 1x = 0$
So $(x+1)(x-1) = x^2 + 1x - 1x -1 =$
$x^2 +(1x-1x) -1 = $
$x^2 + 0 -1 =$
$x^2 -1 $.
Method 1:
$x^2 - 1 = $
$x^2 + x - x - 1 = $
$(x^2 +x) - (x+1) = $
$(x\cdot x + x\cdot 1) + (-1)\cdot(x+1) =$
$x\cdot(x + 1) + (-1)\cdot (x+1) =$
$\color{blue}x\cdot\color{red}{(x + 1)} + \color{blue}{(-1)}\color{red}{(x + 1)} = $
$(\color{blue}x + \color{blue}{(-1)})\color{red}{(x+1)} = $
$(\color{blue}x-\color{blue}1)\color{red}{(x+1)}$
Method 2:
$(x^2 - 1)\div (x+1) = ?????$
We need to find a firs term, $a$ so that when we multiply $a$ times $x+1$ and get $ax+a$ that $ax = x^2$. What can be $a$?  Well $ax=x^2$ and so (if we assume $x$ isn't always $0$) then $a= x$.
So $x(x+1) = x^2 + x$.
But $x^2 -1 \ne x^2 + x$ we must find a "remainder".
$(x^2 - 1)- (x^2 + x) = (x^2-x^2) + (-1-x) = -x - 1$.
So $x^2 - 1=  (x^2 + x) - x-1 = x(x+1) - x-1$.
And $(x^2 - 1)\div (x+1) = x + \frac {-x-1}{x+1}$
Now we must divide $x+1$ into $-x-1$.  
We must find term $b$ so that when we multiply $b$ by $x+1$ and get $bx + b$ the first term $bx = -x$.  What can that $b$ be.  Clearly it is $b=-1$.
So $-1(x+1) = -x-1$.
And $-x-1$ DOES equal $-1(x+1)$ so we have no remainder.
So $(x^2 -1) = x(x+1) + (-1)(x+1)$.
And $(x^2 - 1)\div (x+1) = x + (-1) = x-1$
So $(x^2 - 1) \div (x+1) = x-1$ so that mean $(x^2-1)= (x-1)(x+1)$.
Method 3:
$(x^2 -1) = (x+a) (x+b)$
$= x^2 +ax +bx + ab =$
$x^2 + (a+b)x + ab$
And that is supposed to be $x^2 -1 = x^2 + 0*x + (-1)$.
So we need $a,b$ so that $a*b = -1$ and $a+b = 0$.
Okay so $ab = -1$ so $a = -\frac 1b$ 
And $-\frac 1b + b = 0$ so 
$b = \frac 1b$ so 
$b^2 = 1$.
So $b = \pm 1$.
And $a +b = a\pm 1 = 0$ so $a =\mp 1$.
So one of them is $1$ and the other is $-1$.
So $x^2 -1 = (x+a)(x+b) = (x+1)(x-1)$.
Method 4:
If $ax^2 + bx + c = 0$ has two solutions $x= m$ and $x = n$
Then $ax^2 + bx + c = a(x-m)(x+n)$.
So what are the two solutions to $x^2 -1 = 0$
They are:
$x^2 - 1 =0$ so
$x^2 = 1$ so 
$x  = \pm 1$
So $m = 1$ and $n=-1$ are the two solutions and
so $x^2 -1 = (x- 1)(x-(-1)) = (x-1)(x+1)$.
........
But in ALL of these methods the $+1x -1x =0$ and the "cancel out".
You can cancel them to $0$ or pop them out of nowhere from $0$.
The this is you just have to think of it.
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