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Partial Fractions Solving: Denominators having degrees more than 2

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engphy2
#1
Jan28-12, 09:37 PM
P: 6
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)/(x2+a).
But how about denominators like (x3+a) and so on???
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mathman
#2
Jan29-12, 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.
HallsofIvy
#3
Jan29-12, 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.

checkitagain
#4
Jan29-12, 07:07 PM
P: 99
Partial Fractions Solving: Denominators having degrees more than 2

Quote Quote by HallsofIvy View Post
On the other hand, the one you specifically give is relatively simple:
[tex]x^2+ y^2= (x+ y)(x^2+ xy+ y^2)[/tex]
I haven't received any feedback about my issues with the content
in 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]
engphy2
#5
Feb10-12, 11:20 AM
P: 6
so it can only be done to perfect cube terms??
mathman
#6
Feb10-12, 04:31 PM
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Quote Quote by engphy2 View Post
so it can only be done to perfect cube terms??
Not necessarily. But in general you need to find the factors, which may be difficult in practice.
HallsofIvy
#7
Feb10-12, 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.
Mark44
#8
Feb10-12, 08:05 PM
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Quote Quote by checkitagain View Post
I haven't received any feedback about my issues with the content
in 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]
Yes, there were typos in the exponents in HoI's post.


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