Question 349803
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One question per post.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 3)(x\ +\ 3)\ =\ a]


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2x\ -\ 6)(x\ +\ 3)\ =\ 2(x\ -\ 3)(x\ +\ 3)]


by factoring a 2 out of the first binomial.  But since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 3)(x\ +\ 3)\ =\ a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(x\ -\ 3)(x\ +\ 3)\ =\ 2a]


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