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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$\begingroup$ 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. $\endgroup$– Gordon RoyleCommented Feb 3, 2011 at 3:38
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1 Answer
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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 $f(t)$.
answered Feb 2, 2011 at 12:51