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It is well-known that the coefficients of a chromatic polynomial alternate in sign. But is it possible for a chromatic polynomial to have a factor (over $\mathbb{Q}$) with coefficients which do not alternate?

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I have looked quite hard for factors that do not alternate without success... it would be nice if the claim were true but I cannot see any way of approaching it. – Gordon Royle Feb 3 '11 at 3:38

Factor over what field?

Over $\mathbb{R}$, this is false. The roots of chromatic polynomials are dense in the complex plane, a result of Sokal. So, let $f(t)$ be a chromatic polynomial which has a root in the left half plane, $f(-a+bi)=0$. Then $t^2+(2a)t+(a^2+b^2)$ is a factor of $f(t)$.

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Thanks...I meant over $\mathbb{Q}$. – Adam Feb 2 '11 at 13:23

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