Question 465659
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The other tutor's answer is absolutely correct, but it may be helpful at some point to remember that ANY polynomial term can be considered a square.  Hence:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ -\ n]


is the difference of two squares just as much as *[tex \Large a^2\ -\ b^2] is.  You simply have to re-write it as:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sqrt{m}\right)^2\ -\ \left(\sqrt{n}\right)^2]


and then the factorization would be:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sqrt{m}\ +\ \sqrt{n}\right)\left(\sqrt{m}\ -\ \sqrt{n}\right)]


Here's another example:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 1]


can be considered as the difference of two squares if you write it:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ (-1)]


And would then factor as:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ i)(x\ -\ i)]


Where *[tex \Large i] is the imaginary number defined by *[tex \Large i^2\ =\ -1]


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