Factors question

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Hom
Posts: 39
Joined: Sat Oct 01, 2011 3:22 am

Factors question

Post by Hom » Wed Nov 09, 2011 2:29 am

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

PNT
Posts: 37
Joined: Fri Mar 11, 2011 9:01 pm

Re: Factors question

Post by PNT » Wed Nov 09, 2011 3:54 am

No it is not, try long division.

Hom
Posts: 39
Joined: Sat Oct 01, 2011 3:22 am

Re: Factors question

Post by Hom » Wed Nov 09, 2011 5:46 am

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,

deckoff8
Posts: 23
Joined: Tue Nov 08, 2011 11:12 am

Re: Factors question

Post by deckoff8 » Wed Nov 09, 2011 8:56 pm

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.

mhyyh
Posts: 9
Joined: Mon Oct 10, 2011 9:12 pm

Re: Factors question

Post by mhyyh » Wed Nov 09, 2011 9:13 pm

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

PNT
Posts: 37
Joined: Fri Mar 11, 2011 9:01 pm

Re: Factors question

Post by PNT » Wed Nov 09, 2011 10:46 pm

in that case use the euclidean algorithm and you get x+2

mhyyh
Posts: 9
Joined: Mon Oct 10, 2011 9:12 pm

Re: Factors question

Post by mhyyh » Thu Nov 10, 2011 12:13 am

PNT wrote:in that case use the euclidean algorithm and you get x+2
and 1....

Hom
Posts: 39
Joined: Sat Oct 01, 2011 3:22 am

Re: Factors question

Post by Hom » Thu Nov 10, 2011 12:39 am

mhyyh wrote:
PNT wrote:in that case use the euclidean algorithm and you get x+2
and 1....
Thank you guys. Appreciate it.



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