Question 518091
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I can't make heads or tails out of most of your mess.  Use lots of parentheses and a forward slash to indicate the vinculum next time.  And either capitalize all of your variables or don't capitalize any of them.  X and x are two different things.


Also, one problem per post.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{8x^2\ -\ 50y^2}{8x^2\ -\ 18x\ -\ 5y^2}]


Factor a 2 out of the numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2(4x^2\ -\ 25y^2)}{8x^2\ -\ 18x\ -\ 5y^2}]


Factor the difference of two squares in the numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2(2x\ -\ 5y)(2x\ +\ 5y)}{8x^2\ -\ 18x\ -\ 5y^2}]


Factor the denominator:  (Hint: you should expect at least one of your numerator factors to show up in the denominator, otherwise the original fraction is irreduceable)


Try *[tex \Large 2x\ +\ 5y].  That would mean that the first term of the other factor would have to be *[tex \Large 4x], the second term would have to be *[tex \Large y] and because the first factor has a plus sign, and the *[tex \Large y^2] term sign is negative, the second factor must have a minus sign.


So do FOIL on *[tex \Large (2x\ +\ 5y)(4x\ + y)] to see if it equals *[tex \Large 8x^2\ -\ 18xy\ -\ 5y^2].  If it works you are done and can eliminate the *[tex \Large 2x\ +\ 5y] factor from both numerator and denominator.  If it doesn't work then you have to try *[tex \LARGE 2x\ -\ 5y].


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