Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $f(x)$ and $g(x)$ are two polynomials such that the polynomial $h(x)=xf(x^{3})+x^{2}g(x^{6})$ is divisible by $x^{2}+x+1$, then choose the correct option:

$A. f(1)=g(1)$

$B. f(1) $ is not equal to $g(1)$

$C. f(1)$ and $g(1)$ are non-zero, $f(1)=g(1)$

$D. f(1)=-g(1)$, $f(1)$ and $g(1)$ are non-zero.

What I did: Since $x^{2}+x+1$ divides the $h(x)$, $\omega$ and $\omega^{2}$ are roots of $h(x)$. Therefore, $\omega f(1)+\omega^{2}g(1)=0$ and $\omega^{2} f(1)+\omega g(1)=0$ which gives $f(1)=-g(1)$. But the answer is option $A$. Please help...

NOTE: $\omega$ is $e^{\iota 2\pi /3}$, $\iota =(-1)^{1/2}$

share|improve this question
    
No prob. Deleted. –  Mike Miller Dec 15 '13 at 6:41

3 Answers 3

up vote 2 down vote accepted

You're correct that $f(1)=-g(1)$. But, as Ian points out above, we also have that $g(1)=f(1)$. Put these two facts together, and you'll see that the only correct answer is A!

share|improve this answer
    
Thanks... Didn't note that !! –  Apurv Dec 15 '13 at 6:45

EDIT: I think you mean $\omega=(1)^{1/3}$. You're left with $$ \omega f(1)+\omega^2g(1)=\omega^2 f(1)+\omega g(1)\implies (\omega^2-\omega)(g(1)-f(1))=0 $$

share|improve this answer
    
No. $(-1)^{1/3}$ doesn't satisfy $x^2+x+1=0$, it satisfies $x^2-x+1=0$. On the other hand, the manipulations you've got here are correct (and indeed are only true if $\omega = e^{2\pi i/3}$). –  Mike Miller Dec 15 '13 at 6:41
    
No, the minimal polynomial of the third roots of unity is $x^2+x+1=0$. –  Ian Coley Dec 15 '13 at 6:43
    
No.omega is the cube root of unity... –  Apurv Dec 15 '13 at 6:43
    
That's correct. $(-1)^{1/3}$ is not a third root of unity, it's a sixth root of unity. –  Mike Miller Dec 15 '13 at 6:44
    
Mike is right on the money. There is an error in one of the answers above. –  Doc Dec 15 '13 at 6:45

It seems to me that $f(1)$ and $g(1)$ satisfy two equations which are linearly independent. Hence $f(1) = g(1) = 0$. That implies (A), as well as your answer.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.