How Can You Factorize the Polynomial \( x^6 - 1 \)?

[roam]

Homework Statement

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.

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...

 

[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)

 

[roam]

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)

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

 

[Bohrok]

Do you know about http://en.wikipedia.org/wiki/Polynomial_long_division" ? 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)(x^n-1 + x^n-2a + x^n-3a² + ··· + x²a^n-3 + xa^n-2 + a^n-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.

 

[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).

 

[Hurkyl]

x⁶-1 can be factored to:

(x-1)(x⁵+x⁴+x³+x²+x+1)

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

 

[lurflurf]

you can divide by known factors
I would start with
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

 

[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.

 

[Bohrok]

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).

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

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.

Did Halls delete his message?

 

[Gib Z]

My mistake, I meant Hurkyl.

 

[roam]

Hi Bohrok, I already read your links.

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...

 

[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.

 

[Mentallic]

Hi Bohrok, I already read your links.

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...

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 $$z=x^2$$

 

[roam]

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

(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.

 

[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 $$x^6-1$$