Polynomial Factorisation Q (1 Viewer)

[QZP]

Factorise x^15 + 1 into four factors with real coefficients.

Here's what I got:

x^15 + 1 = (x^5)^3 + 1 = (x+1)(x^4-x^3+x^2-x+1)(x^10-x^5+1)

and x^15 + 1 = (x^3)^5 + 1 = (x+1)(x^2-x+1)(x^12-x^9+x^6-x^3+1)

Don't know what to do from here...

 

[Squar3root]

The best I got is (x^5 +1)(x^10 -x^5 +1)

Are you meant to make it look a certain way? I think this can be factorised further over the imaginary field but it's only extension 1

 

[QZP]

The answer is (x+1) (x^2-x+1) (x^4-x^3+x^2-x+1) (x^8+x^7-x^5-x^4-x^3+x+1)

Note the similar elements to the form I expressed above

 

[Chris100]

Find the complex and real roots of x^15=-1, let the roots be z0,z1,z2...z15 and rewrite x^15+1 as (x-z0)(x-z1)...(x-z15)
when you find the roots, each root will have a conjugate pair which you can expand and simplify to form 4 factors with real coefficients

 

[iStudent]

Find the complex and real roots of x^15=-1, let the roots be z0,z1,z2...z15 and rewrite x^15+1 as (x-z0)(x-z1)...(x-z15)
when you find the roots, each root will have a conjugate pair which you can expand and simplify to form 4 factors with real coefficients

That's the ugly 4u method

Using what Q Z P found earlier:
x^15 + 1 = (x^5)^3 + 1 = (x+1)(x^4-x^3+x^2-x+1)(x^10-x^5+1)
x^15 + 1 = (x^3)^5 + 1 = (x+1)(x^2-x+1)(x^12-x^9+x^6-x^3+1)
so, on equating
(x+1)(x^2-x+1)(x^12-x^9+x^6-x^3+1) = (x+1)(x^4-x^3+x^2-x+1)(x^10-x^5+1)
divide both sides by x^2-x+1 and x+1
to get (x^12-x^9+x^6-x^3+1) = (x^4-x^3+x^2-x+1)(x^10-x^5+1)/(x^2-x+1)
Now, (x^10-x^5+1)/(x^2-x+1) = (x^8+x^7-x^5-x^4-x^3+x+1) by using polynomial long division
so(x^12-x^9+x^6-x^3+1) = (x^4-x^3+x^2-x+1)(x^8+x^7-x^5-x^4-x^3+x+1)

subbing this ugly expression for (x^12-x^9+x^6-x^3+1) into
x^15 + 1 = (x+1)(x^2-x+1)(x^12-x^9+x^6-x^3+1)
to get:
(x+1)(x^2-x+1)(x^4-x^3+x^2-x+1)(x^8+x^7-x^5-x^4-x^3+x+1)
Which is the answer

 

[Squar3root]

but its only extension 1so complex numbers probably wouldn't be the way to go (I think they'll still accept it in an exam). Also it seems like a pretty redundant question, I wouldn't think somehthing like this would be asked. It would be more beneficial to do past papers

 

[QZP]

I thought it was a very nice question (I was working with iStudent on skype). Past papers feel more redundant imo; rather have my algebra skills be pushed and break down on harder q's like this that are not 'mainstream'.

 

[Squar3root]

I thought it was a very nice question (I was working with iStudent on skype). Past papers feel more redundant imo; rather have my algebra skills be pushed and break down on harder q's like this that are not 'mainstream'.

Sounds like a good idea so you can develop the thinking behind the questions