Factoring $2^{2x} +2^{-2x}-2$ I have been stumped by this question asking to solve for $x$ in an equation with negative exponents. The combination of positive and negative exponents is what is confusing me so much. 
So, how would one factor this?
$$2^{2x} +2^{-2x}-2$$
I've tried substituting $2^{2x}$ into another variable like $a$ but haven't had any success. Thank you.
 A: Well, you set $a = 2^{-2x}$ and don't substitute you get:
$2^{4x}a + a - 2*2^{2x}a = a(2^{2x} - 1)^2$
Or in other words
$ 2^{2x} +2^{-2x}-2 =$
$2^{-2x}(2^{4x}-2*2^{2x} + 1) =$
$2^{-2x}(2^{2x} - 1)^2$
And we can go further
$= 2^{-2x}(2^x+1)^2(2^x - 1)^2$
If it helps, but it probably won't, $2^x -1 = 1+ 2+4 + ..... +2^{x-1}$
A: Alternative method:
$$\ 2^{2x} +2^{-2x}-2=\\
(2^x)^2+(2^{-x})^2-2=\\
(2^x+2^{-x})^2-4=\\
(2^x+2^{-x}-2)(2^x+2^{-x}+2).$$
A: If you set $2^x=a$ you are trying to factor $a^2-2+a^{-2}$.  Seeing the symmetry should prompt you to thing about $a\pm \frac 1a$
A: What's the issue with using $a = 2^{2x}$? (Obviously it doesn't change anything at all but it makes the expression somewhat easier to handle)
We have $a+\frac1a-2={a^2-2a+1\over a} = {(a-1)^2\over a} = \left(\sqrt a -\frac 1{\sqrt a}\right)^2$
So, our answer is: $$\boxed{(2^x - 2^{-x})^2}$$
A: Suppose $a = 2^x$, $b=2^{-x}$. First we should note that $ab=1$.
Then the equation becomes $$2^{2x}+2^{-2x}-2 = (2^{x})^2 + (2^{-x})^2 - 2 = a^2 + b^2 -2ab = (a-b)^2 = (2^x - 2^{-x})^2$$
Alternatively, we could just write
$$2^{2x}+2^{-2x}-2 = 2^{2x} - 2 + 2^{-2x} = (2^{x})^2 - 2(2^{x}\cdot 2^{-x}) + (2^{-x})^2 = (2^x - 2^{-x})^2$$
Some people may find it helpful to write $2^{2x} - 2 + 2^{-2x}$ as $2^{2x} - 2\cdot 1 + 2^{-2x} = 2^{2x} - 2\cdot 2^{0x} + 2^{-2x}$. This may make it easier to see that you can factor out $2^{-2x}$, as follows:
$$2^{2x} - 2\cdot 2^{0x} + 2^{-2x} = 2^{-2x}(2^{4x} - 2\cdot 2^{2x} + 1) = 2^{-2x}((2^{2x})^2 - 2\cdot 2^{2x} + 1)$$
