Question 486078
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Well, *[tex \Large x\ -\ 5y] is the quotient although it is an incomplete answer because there is a remainder involved. *[tex \Large x] remainder 0 is completely out to lunch.  (*[tex \Large (x)(3x\ -\ 5y)\ =\ 3x^2\ -\ 5xy\ \neq\ 3x^2\ -\ 25y^2])


The reason *[tex \Large \frac{x^2\ -\ 36}{x\ +\ 6}] comes out so tidily is NOT because "x goes in to x-squared x times and 6 goes into 36 six times" but rather because of the factorization of the numerator.  Note that the numerator is the difference of two squares, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2\ -\ 36}{x\ +\ 6}\ =\ \frac{(x\ +\ 6)(x\ -\ 6)}{x\ +\ 6}\ =\ x\ -\ 6,\ x\ \neq\ -6]


Considering the number 3 to be the square of *[tex \Large \sqrt{3}], you can factor the numerator of your problem as the difference of two squares:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2\ -\ 25y^2\ =\ \left(\left(\sqrt{3}\right)x\ -\ 5y\right)\left(\left(\sqrt{3}\right)x\ +\ 5y\right)]


but neither of these factors is evenly divisible by your given denominator, *[tex \Large 3x\ +\ 5y]


The only thing left to us is to do multi-variable polynomial long division.  A process that is very difficult and time consuming to render on this system.  Perhaps you can get an idea from:


<a href="http://www.mathsisfun.com/algebra/polynomials-division-long.html
">MathIsFun Polynomial Long Division</a>  (scroll all the way down to the bottom of the page for the multi-variable process)


Properly performed you should obtain the quotient that you mentioned, but also with a remainder of *[tex \Large 10xy]


In summary:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3x^2\ -\ 25y^2}{3x\ +\ 5y}\ =\ x\ -\ 5y\ +\ \frac{10xy}{3x\ +\ 5y}] 


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