Tell me what it would look like to factor $x^2-1 = (x+1)(x-1)$ a different way

Question

During the factoring of $x^2-1$ I saw a $+x$ and $-x$ were introduced but I wonder how the factoring would go if the $+x-x$ were added in reverse, like so $-x+x$.

I was shown $x^2-1$ can be factored to $(x+1)(x-1)$ thusly...
$x^2-1 =$
$x^2+x-x-1 =$
$(x^2+x)-(x+1) =$
$(x*x+x*1)+(-1)*(x+1) =$
$x*(x+1)+(-1)*(x+1) =$
$(x+(-1))(x+1) =$
$(x-1)(x+1) =$

I want to know how $x^2-1$ can be factored to $(x+1)(x-1)$ if instead of $x^2+x-x-1 =$ the $+x$ and $-x$ were brought in the other way around like so $x^2-x+x-1 =$

I tried to factor this to $(x-1)(x+1)$ and got lost along the way.

I'll start the equation again.
$x^2-1 =$
$x^2-x+x-1 =$
... what happens next?

Answer 1

I don't understand what you don't understand, First of all, Addition is a commutative operation so, it won't matter how you "add" You're lost at this point..? $x^2-x+x-1$ Hint: Just Take the common factor out, like you did first, earlier you took $+1$ common. Something else this time. and obviously, $+x$ from the first two terms.

Just remember that the result would be the same, however you factor out.

If you still couldn't understand, $$x^2-x+x-1$$ $$=(x)(x-1)+(+1)(x-1)$$ $$ (x-1)[(x)+(+1)]$$ $$\implies (x-1)(x+1)$$

Answer 2

$$x^2-x+x-1=x*x-x*1+(x-1)$$or

$$x(x-1)+(x-1)$$ Which gives you what you want