# Polynomials: Linear Factors

1. Sep 23, 2009

### roam

1. The problem statement, all variables and given/known data
Consider the polynomial $$p(x)=x^6-1$$. (Apply over any field $$F$$).

(a) Find two elements $$a,b \in F$$ so that $$p(a)=p(b)=0$$. Then use your answer to find two linear factors of $$p(x)$$.

(b) Show that the other factor of $$p(x)$$ is $$x^4+x^2+1$$

(c) Verify the identity $$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$$ and hence factor $$p(x)$$ as a product of two linear factors and two quadratic factors.

3. The attempt at a solution

(a) $$x^6-1=0$$ can be re-written as $$x.x^5-1=0$$, therefore x=1 or -1. ±1 are the two roots of the équation. So I guess two linear factors would be $$(x+1)$$ and $$(x-1)$$. Is this correct?

(b) I'm not quite sure how to show this one because I can't figure out what the question wants us to show...

2. Sep 23, 2009

### Bohrok

You know that (x-1) and (x+1) are factors. Now show that the other factor is x4 + x2 + 1, or that p(x) can be factored as (x-1)(x+1)(x4 + x2 + 1)

3. Sep 24, 2009

### roam

That's the part I'm stuck on... I don't see how it can be factored like that.

4. Sep 24, 2009

### Bohrok

Do you know about http://en.wikipedia.org/wiki/Polynomial_long_division" [Broken]? It also helps to know that x6 - 1 can be factored following this pattern:
x2 - a2 = (x - a)(x + a)
x3 - a3 = (x -a )(x2 + xa + a2)
x4 - a4 = (x - a)(x3 + x2a + xa2 + a3)

The generalization is:
xn - an = (x - a)(xn-1 + xn-2a + xn-3a2 + ··· + x2an-3 + xan-2 + an-1)

There's a better way of finding -1 and +1 as the roots of p(x):
x6 - 1 = (x3)2 - 1 = (x3 - 1)(x3 + 1)
then let it equal 0 and solve for x.

Last edited by a moderator: May 4, 2017
5. Sep 25, 2009

### roam

Thanks. So p(x)=x6-1 can be factored to:

(x-1)(x5+x4+x3+x2+x+1)

Now I still can't see how to show that $$x^4+x^2+1$$ is the other factor of p(x).

6. Sep 25, 2009

### Hurkyl

Staff Emeritus
How do you know this is correct? How might you go about proving that x6-1 is really equal to this product?

7. Sep 25, 2009

### lurflurf

you can divide by known factors
x^6-1=(x^3+1)(x^3-1)
and
x^3-a^3=(x-a)(x^2+ax+a^2)
a=1 and -1 cases

8. Sep 25, 2009

### Gib Z

I think the easiest way to solve the part the OP wants is what Halls was hinting towards to. Read the question carefully, its not asking you to derive or prove, its asking for something simpler than that.

9. Sep 25, 2009

### Bohrok

Now use polynomial division or synthetic division. Look at the links in my last post.

Did Halls delete his message?

10. Sep 25, 2009

### Gib Z

My mistake, I meant Hurkyl.

11. Sep 26, 2009

### roam

I divided the polynomial p(x)=x6-1 by x4+x2+1 and the quotient was x2-1, no remainder. So it is a factor of p(x). Is this all I needed to show?

Now can you help me get started with part (c) please, I've no idea...

12. Sep 26, 2009

### Bohrok

What you should do first is use (x-1) and (x+a), since you knew they were zeros, to divide x6 -1 by, then you'd get x4 + x2 + 1. This way you show the other factor is x4 + x2 + 1.

13. Sep 26, 2009

### Mentallic

This is kind of backwards. You found the two linear factors $$(x+1)(x-1)$$ so divide $$x^6-1$$ by $$(x+1)(x-1)$$ or $$x^2-1$$ and you'll end up with $$x^4+x^2+1$$.

Also notice it's equivalent to factorizing the difference of 2 cubes:
$$z^3-a^3=(z-a)(z^2+az+1)$$
where $$z=x^2$$

14. Sep 28, 2009

### roam

I understand (b) now! But not quite sure how to go about doing part (c):

15. Sep 28, 2009

### Mentallic

Well, expand the RHS and show it equals the LHS.
Then, since you've found from b) that: $$x^6-1=(x-1)(x+1)(x^4+x^2+1)$$
and you've verified from the first part in c) that the quartic factor $$x^4+x^2+1$$ is equal to those 2 quadratic factors, once you show the quadratic factors cannot be simplified by showing they have complex roots or otherwise, hence completely factorize $$x^6-1$$