Partial Fractions Solving: Denominators having degrees more than 2by engphy2 Tags: degrees, denominators, fractions, partial, solving 

#1
Jan2812, 09:37 PM

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I just want to know if there is now a way to solve fractions like which had a variables that has a degree more than 2 in its denominator. I know that denominators having degrees of 2 could be solved using
(Ax + B)/(x^{2}+a). But how about denominators like (x^{3}+a) and so on??? 



#2
Jan2912, 03:55 PM

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The main difficulty is that it is much harder to get the factors for higher degree polynomials. The principal is the same once you've got the factors.




#3
Jan2912, 04:24 PM

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On the other hand, the one you specifically give is relatively simple:
As checkitagain pointed out, this should have been [tex]x^2+ y^2= (x+ y)(x^2 xy+ y^2)[/tex] By "completing the square", [itex]x^2 xy+ y^2= x^2 xy+ y^2/4 y^2/4+ y^2= (x y/2)^2+ 3y^2/4[/itex]. Let [itex]y= \sqrt{a}[/itex]. But, as mathman says, while every polynomial can be factored into linear or quadratic terms over the real numbers (into linear terms over the complex numbers), the higher the degree of the polynomial, the harder to find the factors. 



#4
Jan2912, 07:07 PM

P: 99

Partial Fractions Solving: Denominators having degrees more than 2in the quote box above, so I am posting this: [tex]x^3  y^3 = (x  y)(x^2 + xy + y^2)[/tex] [tex]x^3 + y^3 = (x + y)(x^2  xy + y^2)[/tex] 



#5
Feb1012, 11:20 AM

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so it can only be done to perfect cube terms??




#6
Feb1012, 04:31 PM

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#7
Feb1012, 05:24 PM

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Every polynomial of degree greater than two can be factored into first or second degree factors in fact, if we use complex numbers entirely into first degree factors. That does NOT mean that there is any simple way to find those factors.




#8
Feb1012, 08:05 PM

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