Factorisation of H-palindromic polynomial Background
From what I understand, a normal palindromic polynomial of degree $2m$ with real valued coefficients:
$$p(x) = \sum_{j=0}^{2m} a_j x^j$$
where $ a_j = a_{2m-j}$, e.g. $x^4 + 2x^3 + 3x^2 + 2x+ 1$
can be represented as a degree $m$ polynomial
$$q(x) = x^m\sum_{j=0}^{m} b_j X^j$$
where $X = (x + x^{-1})$. The coefficients $b_j$ can be determined from $a_j$ without having to find the roots. The good thing about this is that the order of the polynomial is reduced by half and thus it is easier to find the roots.
http://en.wikipedia.org/wiki/Palindromic_polynomial
Question
Can something similar be done with H-palindromic polynomials, to help with finding the roots? Edit: $x \in \mathbb{C}$, specifically $x = e^{i \theta}$. 
These are:
$$r(x) = \sum_{j=0}^{2m} c_j x^j$$
where $c_j \in \mathbb{C}, c_j = \bar{c}_{2m-j}$.
e.g. $(1+1i)x^4 + (2+3i)x^3 + 3x^2 + (2-3i)x+ (1-1i)$
I have read that they can be written:
$$r(x) = \prod_{j=1}^{m}(\alpha_j x^2 + \beta_j x + \bar{\alpha_j})$$
where $\alpha_j \in \mathbb{C}, \beta_j \in \mathbb{R}$, but I'm not sure if this helps.
Supplementary
Are there any root-finding algorithms that make use of this symmetry when finding the roots?
 A: First there are numerical methods that exploit these types of symmetries. I don't know the detail so it is better to wait for other answers or read the papers that appear when you google the keywords numerical roots palindromic polynomial.
An observation that can be made, but that I don't know if it moves in the right direction, is the following:
If a polynomial is H-palindromic then $$\overline{P(\overline{x})}=x^{d}P\left(\frac{1}{x}\right)$$
where $d:=\text{deg}(P)$.
Therefore, for all $x\in\mathbb{R}$
 $$|P(x)|^2=P(x)\overline{P(x)}=P(x)\overline{P(\overline{x})}=x^dP(x)P\left(\frac{1}{x}\right)$$
This means that $Q(x):=x^dP(x)P\left(\frac{1}{x}\right)$ is a polynomial that returns real numbers for all real numbers (hence it has real coefficients) and satisfies 
$$x^{2d}Q\left(\frac{1}{x}\right)=x^{2d}\frac{1}{x^d}P\left(\frac{1}{x}\right)P(x)=Q(x).$$
Hence $Q$ is palindromic and its roots are the roots of $P$ and their reciprocals. 
On one hand the degree of $Q$ doubled, but now its coefficients are real. Moreover, using the trick for palindromic polynomials finding its roots reduces to finding the roots of a polynomial $R(x)$ of degree $d$ again such that $Q(x)=x^dR\left(x+\frac{1}{x}\right)$, but with real coefficients. 
The advantage of real coefficients is that now you can find the roots (whether numerically or exactly) staying in the real numbers.
