# Partial fractions question

1. Sep 23, 2007

### rootX

It says in my book that
a any function can be decomposed to some sum of strictly proper rational functions where the denominator of each rational function is either consist of linear functions, irreducible quadratic functions.

"Any proper rational function can be expressed as a sum of simpler rational functions whose den's are either linear functions or irreducible quadratic functions." [here's the exact wording]

I was thinking what happens when the denominator has a irreducible 4th degree factor?

2. Sep 23, 2007

### arildno

You do not have irreducible 4th degree polynomials; every polynomial with real coefficients can be factorized in terms of linear and quadratic polynomials with real coefficients.

A fourth-degree real polynomial whose roots are all complex can be decomposed into two irreducible real quadratic polynomial factors.

For example,
$$x^{4}+1=(x^{2}+\sqrt{2}x+1)(x^{2}-\sqrt{2}x+1)$$

Last edited: Sep 23, 2007
3. Sep 23, 2007

### rootX

Thanks.

But is there a way to decompose 4th powers in to quadratics?
I saw that example yesterday in my book, but it does not mention any strategy to do this.
I don't see any ><.

4. Sep 23, 2007

### arildno

There is no foolproof manner of doing this, unless you delve into the cumbersome general solution of a fourth-degree equation.

That is NOT a simple procedure!

5. Sep 23, 2007

### HallsofIvy

Staff Emeritus
Normally, "irreducible" means it cannot be factored into lower degree polynomials with integer coefficients. Any polynomial can be factored into linear factors over the complex numbers (the complex numbers are "algebraically complete"). Since any complex zeroes of a polynomial with real coefficients come in "conjugate" pairs, the pairs can be multiplied to give a quadratic factor with real, not necessarily rational or integer, coefficients. You notice that the example arildno gave had a $\sqrt{2}$ coefficient.

Other than "manually" factoring (very difficult if your polynomial has irrational roots) or otherwise finding the zeroes of the polynomial, I don't believe there is any simple way of factoring polynomials of large order. (And, of course, polynomials of order higher than 4 may have zeroes that cannot be written in terms of roots!)

Last edited: Sep 23, 2007