# Factoring a Root of Unity

1. Jan 22, 2012

### spongegar

I need to factor x5 - 1.

I know (x-1) is a factor and have gotten:
(x-1)(x4+x3+x2+x +1)

I'm not sure where to go from here.

2. Jan 22, 2012

### Mentallic

So what you're looking for is to solve $x^5-1=0$

A better approach to solving this is to convert 1 into its mod-arg form.

So we have $x^5=1$ and $1=cis(2\pi)$ or $e^{2i\pi}$ if you're more accustomed to that. Now can you find the roots of unity?

3. Jan 22, 2012

### spongegar

Yes, I know that is a much more convenient approach and have used that to find the roots, but I need to factorize it for my assignment.
Thanks.

4. Jan 22, 2012

### Mentallic

Finding the roots of unity is a very handy method of factorizing that quartic $x^4+x^3+x^2+x+1$ into two irreducible quadratics.

Notice that $$(x-\alpha)(x-\bar{\alpha})=x^2-(\alpha+\bar{\alpha})x+\alpha\bar{\alpha}$$

where $\alpha$ is a complex root and $\bar{\alpha}$ is the conjugate root. If you let $\alpha=cis\theta$ then you can show that $\alpha+\bar{\alpha}$ and $\alpha\bar{\alpha}$ are both real values.

5. Jan 22, 2012

### A. Bahat

A complex number $a$ is a root of the polynomial $P(x)$ if and only $(x-a)$ is a factor of $P(x)$. If you find all of the roots (which are just the 5th roots of unity), then you automatically get the factorization as well.

6. Jan 23, 2012

### Staff: Mentor

If you are just wanting to factorize that quartic, you can input the coefficients here to see the solutions.

7. Jan 23, 2012

### Mentallic

There is a much better way of finding the roots than using that site which merely gives you numerical approximations.

8. Jan 24, 2012

### Staff: Mentor

Numerical approximations may be all that OP needs. He hasn't indicated.

9. Jan 24, 2012

### Mentallic

Yes he has

10. Jan 24, 2012

### Staff: Mentor

The roots will give him the factors. 10 sig figs may be sufficient for OP's needs. He hasn't indicated.

11. Jan 24, 2012

### ehild

The fifth roots of 1 are ak=exp(2pi/5*k). Following Mentallic's suggestion, the second-order factors are

x^2-2xcos(72°)+1 and x^2 - 2 xcos(144°)-1.

It can be derived from the regular pentagon that

$$\cos(72°)=\frac{\sqrt5-1}{4}$$

and $$\cos(144°)=-\frac{\sqrt5+1}{4}$$.

With these, the exact expression of the factors are obtained.

ehild

Last edited: Jan 24, 2012
12. Jan 24, 2012

### Mentallic

He may not have specifically said "I need to find the roots in exact form to factorize x5-1" but when you factor something, you don't do it with numerical approximations, you do it with exact roots - especially when the roots are relatively easy to find and it's a common question to be asked when studying complex numbers.

I've seen numerical approximations to equations, but never have I seen them taken a step further and put into a factorized form.

For example, I wouldn't be satisfied with $$x^2-2\approx(x-1.414)(x+1.414)$$ when $$(x-\sqrt{2})(x+\sqrt{2})$$ is readily available. And I've never seen numerical roots being applied in such a fashion either, but it might be used like that in some cases I guess.

13. Jan 24, 2012

### Staff: Mentor

That's right, he didn't.
If the numerical approximations are all that's needed, then they may well suffice.
I think you might be satisfied with the 1.414 approximation, sometimes.

OP indicated he needed "to factorize it for my assignment". Sure, that task may constitute the assignment, in which case the exact solution would be required. But maybe his assignment was a curve-sketching task with no marks for the exact solution, or the factorizing was incidental to a physics assignment where the numerical approximation would be what was used anyway. Only the OP can know, and he's not saying. I didn't seek to restrict his options to exact solutions only.

In any case, ready access to approximations affords an easy way for him to check his exact solution.

14. Jan 25, 2012

### Mentallic

If you can show me one example of where numerical approximations to roots of polynomials have been put in factored form, then you have a case. Until then, it just looks to me as though you're desperately grabbing at straws to try back up what you initially said.

I'm not here to point the finger and laugh at your honest mistake, I'm only trying to set the facts straight that if an assignment asks for a polynomial to be factored, then it must be in exact form. If it asked for the solutions however, then they may be numerical approximations.

15. Jan 25, 2012

### Staff: Mentor

I made no mistake. You assumed he needed exact solutions, and that's fair enough. But I made no such assumption. OP's wording had a nuance that hinted his need may be part of an assignment, not the totality of it. Were this to be a case where he'd end up using the numerical values anyway, he may not even need to spend time evaluating the exact factors. I pointed out his options were open; OP can make the judgement on what suits his need.