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Factors question

Posted: Wed Nov 09, 2011 2:29 am
by Hom
Is X^2+X+1 a factor of X^3+X+1 in Z/3Z?

I was thinking since X=1 make X^2+X+1=0 and also X^3+X+1, can we say that there exists a P(X) that X^3+X+1 = (X^2+X+1) p(X)

Thanks

Re: Factors question

Posted: Wed Nov 09, 2011 3:54 am
by PNT
No it is not, try long division.

Re: Factors question

Posted: Wed Nov 09, 2011 5:46 am
by Hom
PNT wrote:No it is not, try long division.
I tried to use this method,
Assuming (x^2+x+1)(ax+b) = ax^3+(a+b)x^2+(a+b)x+(b+1) = X^3+X+1,
then by checking X^3 and the constant, we got a=3n+1 , b = 3m (n,m are of Z). then (a+b)X^2 --> X^2 !=0 which contradicts the assumption.

is this logic ok or any better solution to check?

Thanks,

Re: Factors question

Posted: Wed Nov 09, 2011 8:56 pm
by deckoff8
well since x^2+x+1 and x^3+x+1 both have 1 as the leading coeff and the zero power coeff you know ax+b has to have the form x+1, which doesnt work.

Re: Factors question

Posted: Wed Nov 09, 2011 9:13 pm
by mhyyh
Well, I think Hom just misunderstood the question and led us to another direction....
I have his materials, and i think the question is:
What are the common divisors of x^2+x+1 and x^3+x+1 in Z/3Z

Re: Factors question

Posted: Wed Nov 09, 2011 10:46 pm
by PNT
in that case use the euclidean algorithm and you get x+2

Re: Factors question

Posted: Thu Nov 10, 2011 12:13 am
by mhyyh
PNT wrote:in that case use the euclidean algorithm and you get x+2
and 1....

Re: Factors question

Posted: Thu Nov 10, 2011 12:39 am
by Hom
mhyyh wrote:
PNT wrote:in that case use the euclidean algorithm and you get x+2
and 1....
Thank you guys. Appreciate it.