Question 520369
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I think that you are saying you have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6(t\ +\ w)^2\ +\ 11(t\ +\ w)\ -\ 10]


And you want to factor it.


What I would do, and this is just me, would make a simple substitution so that the quadratic doesn't look quite so ugly.  Like this:


Let *[tex \Large u\ =\ t\ +\ w], then substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6u^2\ +\ 11u\ -\ 10]


Which factors rather tidily to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3u\ -\ 2)(2u\ +\ 5)]


Then substitute back:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(3(t\ +\ w)\ -\ 2\right)\left(2(t\ +\ w)\ +\ 5\right)]


Then distribute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(3t\ +\ 3w\ -\ 2\right)\left(2t\ +\ 2w\ +\ 5\right)]


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