Question 459821
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It has the pattern:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ a^2\ -\ b^2]


In fact, neither of the terms absolutely has to be a perfect square.  For example,


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x^2\ -\ 2]


can  be factored as the difference of two squares if you simply recall that 2 is nothing more than the square root of 2 squared, hence:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ x^2\ -\ 2\ =\ \left(x\ +\ \sqrt{2}\right)\left(x\ -\ \sqrt{2}\right)]


So any two term polynomial where the signs on the terms are opposite is, indeed, the difference of two squares.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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