# Solving a Simple nth Order Polynomial

1. Apr 10, 2007

### dimensionless

1. The problem statement, all variables and given/known data
I'm trying to solve the following equation
$$C + x + x^{3} + x^{5} + x^{7} + x^{9} = 0$$

3. The attempt at a solution
I can factor out $$x+B$$, but the factoring process will be very tedious. A little too tedious if I'm barking up the wrong tree. Any suggestions?

2. Apr 10, 2007

### happyg1

I don't think x+B (I think you mean x+C) factors out of that.

CC

3. Apr 10, 2007

### Staff: Mentor

Multiply by $x^2-1$

4. Apr 20, 2007

### dimensionless

I could have B = C

I multiplied by $$x^{2}-1$$ and I got
$$-C-x-Cx^{2}+x^{11} = 0$$

I good improvement, but still not a solution.

Last edited: Apr 20, 2007
5. Apr 21, 2007

### christianjb

I don't see how this can be done analytically. I would use a numerical method.

6. Apr 21, 2007

### d_leet

That can't be correct. If we call the original function p(x) then if we start with

p(x)=0

and multiply by x2-1 then you end up with

(x2-1)p(x)=0

It should be obvious that 1, and -1 are solutions of this equation, but you should notice that they are not solutions of the result you found after multiplying, thus you must have made a mistake somewhere.

7. Apr 22, 2007

### dimensionless

I reworked the multiplication and got a very similar answer:
$$-C-x+Cx^{2}+x^{11} = 0$$

Maybe there is some 11th order 'quadratic' I can use to solve this.

8. Apr 22, 2007

### dimensionless

Yeah, I'm having trouble with it. I do expect, however, that an analytic solution can be found.

9. Apr 22, 2007

### matt grime

I very much doubt your expectation is true. Analytic solutions (as in sensible functions of the coeffecients) probably cannot be found. They do not, in general, exist for any polynomial of degree 5 or more. (Though there are methods for quintics.)

There is, of course, Galois theory. If you could work out the Galois group you might have a chance.

10. Apr 23, 2007

### dimensionless

I imagine that there is an algorithm that can be worked out by hand and not be tooo tedious. Mathematica seems to have a function that finds the roots of a polynomial exactly.

http://documents.wolfram.com/mathematica/functions/Roots

At any rate, I don't see any reason why this shouldn't be possible.

11. Apr 23, 2007

### HallsofIvy

Staff Emeritus
But I'm sure that function does not find roots in terms of an unknown parameter C! That's the problem. Roots and methods of soltution depend strongly on what C is. For one thing, any rational roots of this equation must be integers that evenly divide C.

12. Apr 23, 2007

### matt grime

Sorry, but that is nonsense. It may have functions to find roots to arbitrary precision, but that is not the same as finding the roots. And it won't even have that since a computer has a machine error to contend with.

apart from lots of people with maths phds telling you it can't be done? See, you've made me make proof be appeal to authority! Gah, what a rubbish answer. Let me expand on what I wrote before:

Google Galois Theory. I didn't put those words in at random.

It can be shown that there is no method that allows you to work out the roots of arbitrary polynomials by field operations and n'th roots of the coefficients for deg >4, and for deg 5 there exist methods by elliptic functions. BUT THAT IS IT. Any attempt at a solution by tactics akin to the quadratic formula is DOOMED to fail for arbitrary polys of degree >4.

Last edited: Apr 23, 2007
13. Jan 10, 2009

### dimensionless

This question is not aimed specifically at Matt, but I've done some more reading about this. Everything I seem to find says that polynomials of deg >4 are not solvable by using =, -, *, /, and ^, although Matt has very strongly implied that there are methods for deg 5 polynomials. I still wonder if there may be solutions by other means. What about the use of transforms and operator functions? I'm not totally convinced that it is possible to proove that a solution does not exist by means of transforms, operator functions, and other methods.