<s>[INST] 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.
 [/INST] 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:
$$2^{2x} - 2\cdot 2^{0x} + 2^{-2x} = 2^{-2x}(2^{4x} - 2\cdot 2^{2x} + 1) = 2^{-2x}((2^{2x})^2 - 2\cdot 2^{2x} + 1)$$
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