Why do parentheses make - a + when $x^2+x-x-1$ is changed to $(x^2+x)-(x+1)$? I know that the equations are equivalent by doing the math with the same value for x, but I don't understand the rules for changing orders or operations.
When it is not the first addition or subtraction happening in the equation, parentheses make the addition subtraction and vice versa? Are there any other rules?
$x^2+x-x-1 = (x^2+x)-(x+1)$?  
What if you put the parentheses around the two $x$s in the middle?
$x^2+(x-x)+1$? Should that be (x+x) in the middle?
 A: Think of subtracting $(x+1)$ as adding $-1\cdot(x+1)$.  
Then apply the distributive property, so it's adding $-1\cdot x + -1\cdot 1=-x-1$.
When you add $(x-x)$, it should be $(x-x)$, not $x+x$, because there's no $-1$ to distribute.
A: The sign before the parenthesis distributes to the content:
$$\begin{align}+(a+b)&=a+b\\-(a+b)&=-a-b\\+(a-b)&=a-b\\-(a-b)&=-a+b.\end{align}$$
A: That's a simple general rule for bracket notation: $-(a+b)=-a-b$. It's more or less a definition
A: Let's simplify it, and suppose you have $x^2 - x - 1$.
Examine what you are doing. You start with $x^2$.
Then you subtract $x$ from $x^2$.
Then you subtract $1$ from what you have got after the previous step, that is, $x^2 - x$.
So you have subtracted first $x$, then $1$, from $x^2$.
What have you in total removed from $x^2$?  You have removed $x$ and $1$, that is, $x+1$, from $x^2$.
So subtracting $x$ and then $1$ from $x^2$ is exactly the same as subtracting $x + 1$ from $x^2$.
There you see, $x^2 - x - 1 \equiv x^2 - (x + 1)$.
You can do it generally, and see that $a - b - c \equiv a - (b + c)$, whatever $a$, $b$ and $c$ are.
