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

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

• "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. – Renoldus Jun 2 '20 at 17:43
• 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. – UmbQbify Jun 2 '20 at 17:47
• 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 $(x-1)(x+1)$ – Renoldus Jun 2 '20 at 17:50
• Fine. I'll edit. – UmbQbify Jun 2 '20 at 17:55
• 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. – Renoldus Jun 2 '20 at 18:02

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

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