# Partial Fractions Solving: Denominators having degrees more than 2

1. Jan 28, 2012

### engphy2

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. Jan 29, 2012

### mathman

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. Jan 29, 2012

### HallsofIvy

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. Jan 29, 2012

### checkitagain

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. Feb 10, 2012

### engphy2

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

6. Feb 10, 2012

### mathman

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

7. Feb 10, 2012

### HallsofIvy

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.

8. Feb 10, 2012

### Staff: Mentor

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