Question 1164306
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 7xy\ +\ 3y^2\ -\ 5y\ -\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 7xy\ +\ (3y\ +\ 1)(y\ -\ 2)]


Note that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7xy\ =\ x(y\ -\ 2)\ +\ 2x(3y\ +\ 1)]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ x(y\ -\ 2)\ +\ 2x(3y\ +\ 1)\ +\ (3y\ +\ 1)(y\ -\ 2)]


Factor out *[tex \Large 3y\ +\ 1]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ x(y\ -\ 2)\ +\ (3y\ +\ 1)(2x\ +\ y\ -\ 2)]


Factor out *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(2x\ +\ y\ -\ 2)\ +\ (3y\ +\ 1)(2x\ +\ y\ -\ 2)]


Factor out *[tex \Large 2x\ +\ y\ -\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 3y\ +\ 1)(2x\ +\ y\ -\ 2)]


Go have a snack.

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