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> Prove the following are true in $A[x]$ for any ring A: For any odd $n$,
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> 1. $x+1$ is a factor of $x^n+1$
> 2. $x+1$ is a factor of $x^n+x^{n-1}+...+x+1$.
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So I believe we do long division and get that $x^n+1=(x+1)(x^{n-1}+...+x+1)$.I don't know that my math is correct here but then this would show $x+1$ is a factor. Then for 2, $x^n+x^{n-1}+...+x+1= (x^n)+(x^{n-1}+...+x+1)$. I think I would do long division again but I'm not sure with what pieces.

Am I on the right track?