Partial Fractions Solving: Denominators having degrees more than 2

1. engphy2

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???

2. mathman

6,462
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. HallsofIvy

40,368
Staff Emeritus
On the other hand, the one you specifically give is relatively simple:
As checkitagain pointed out, this should have been
$$x^2+ y^2= (x+ y)(x^2- xy+ y^2)$$

By "completing the square", $x^2- xy+ y^2= x^2- xy+ y^2/4- y^2/4+ y^2= (x- y/2)^2+ 3y^2/4$.

Let $y= \sqrt{a}$.

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.

Last edited: Jan 30, 2012
4. checkitagain

99
in the quote box above, so I am posting this:

$$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$$

$$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$

5. engphy2

6
so it can only be done to perfect cube terms??

6. mathman

6,462
Not necessarily. But in general you need to find the factors, which may be difficult in practice.

7. HallsofIvy

40,368
Staff Emeritus
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.

Staff: Mentor

Yes, there were typos in the exponents in HoI's post.