One more comment on this frequently answered question: Let $a_0 x^{2n} + a_1 x^{2n-1} + \cdots + a_{n} x^{n} + \cdots a_1 x + a_0$ be a palindromic polynomial with real coefficients and $2k$ isolated roots on the unit circle. Then any sufficiently small perturbation of this polynomial to another real palindromic polynomial also has $2k$ roots on the unit circle.
Proof: Notice that $e^{\pm i \theta}$ is a root of this polynomial if and only if $a_0 \cos (n \theta) + a_1 \cos ((n-1) \theta) + \cdots + a_{n/2-1} \cos \theta + a_0/2=0$. Write $f(\theta)$ for the right hand side of this equation. Our hypothesis is that $f$ has $k$ isolated roots, $\theta_1$, $\theta_2$, ..., $\theta_k$. Then we can find $\epsilon>0$ and disjoint intervals $(a_i, b_i)$ around each $\theta_i$ such that (1) We have $f((a_i, b_i)) \supseteq (-\epsilon, \epsilon)$. (2) On $(a_i, b_i)$, we have $|f'|<\epsilon$ (3) Off of the $(a_i, b_i)$, we have $|f|>\epsilon/2$.
Then, if our perturbation is small enough that $f$ and $f'$ change by less than $\epsilon/4$ everywhere, then the perturbed $f$ still has one root in each $(a_i, b_i)$, and no roots elsewhere. QED
Why do I point this out? Take any of the above examples and perturb its coefficients slightly, while keeping them rational and palindromic. Then you get another example! This observation destroys most attempts to classify such polynomials by number theoretic criteria.
