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?


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, $$x^2-x+x-1$$ $$=(x)(x-1)+(+1)(x-1)$$ $$ (x-1)[(x)+(+1)]$$ $$\implies (x-1)(x+1)$$

  • $\begingroup$ "I don't understand what you don't understand" I'll add to the OP and show my work and how it doesn't get me to the same result. $\endgroup$ – Renoldus Jun 2 '20 at 17:43
  • $\begingroup$ It gets you to the same result , man (>ლ). By that, I meant I don't understand what you possibly couldn't understand in the simple equation where you have to take the common factor out. $\endgroup$ – UmbQbify Jun 2 '20 at 17:47
  • $\begingroup$ I'm sure it gets the same result. I'm learning how to factor equations. I don't know the steps to get from $x^2-x+x-1$ to $(x-1)(x+1)$ $\endgroup$ – Renoldus Jun 2 '20 at 17:50
  • $\begingroup$ Fine. I'll edit. $\endgroup$ – UmbQbify Jun 2 '20 at 17:55
  • $\begingroup$ Ah. I see. I got up to the point where you put a bracket around the parenthesis. I didn't know you could do that. Thank you. $\endgroup$ – Renoldus Jun 2 '20 at 18:02


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


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