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Why must $x^2 + x + 1$ be a factor of $x^5+x^4+x^3+x^2+x+1$?

I know that when we divide $x^5+x^4+x^3+x^2+x+1$ by $x^3+1$ we get $x^2 + x + 1$, but is there an argument/theorem or anything that could tell that $x^2+x+1$ must divide $x^2 + x + 1$?

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  • $\begingroup$ Must divide $x^5+x^4+x^3+x^2+x+1$ by $x^3+1$ instead of $x^2+x+1$? $\endgroup$
    – Mikasa
    Sep 19 '12 at 6:39
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It depends on what you count as telling. It follows from an easy, basic manipulation of familiar identities, but I don’t see any way to recognize this without appealing to something at least somewhat computational. I look at the sum $x^5+x^4+x^3+x^2+x+1$ and immediately think of $x^6-1$, and from there it all falls out:

$$\begin{align*} x^5+x^4+x^3+x^2+x+1&=\frac{x^6-1}{x-1}\\ &=\frac{(x^3)^2-1}{x-1}\\ &=\frac{(x^3-1)(x^3+1)}{x-1}\\ &=(x^3+1)\frac{x^3-1}{x-1}\\ &=(x^3+1)(x^2+x+1)\;, \end{align*}$$

all using standard factorizations.

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  • $\begingroup$ Is Division Algorithm for F[x] makes the OP sure for dividing? He wanted to know that. $\endgroup$
    – Mikasa
    Sep 19 '12 at 6:48
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$(x^2+x+1)(x-1)=x^3-1$ and $(x^5+x^4+x^3+x^2+x+1)(x-1)=x^6-1$. Then note that $x^6-1=(x^3-1)(x^3+1)$. Thus in effect it is the third binmoial formula.

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