Polynomials with symmetric coeffficients Let $f(x)=a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$ where $a_i=a_{n-i}$ for $i=0, \ldots, n-1$. Are there any known properties of such polynomials (such as its properties or a method to compute its roots)?
 A: Well, if $x$ is a root, so is $\frac1x$.  This means if there are an odd number of distinct real roots (say it is an odd degree polynomial), then $\pm1$ is a root.  
Further, if the polynomial is even degree (or after division by $x\pm1$ as mentioned above), you can use the substitution $y = x + \frac1x$ to reduce the degree by a factor of $2$.
A: Let's take a look at an example: $P(x) = x^6 + 4x^5 - 11x^4 + 7x^3 - 11x^2 + 4x + 1$. This polynomial satisfies the conditions of the coefficients that: $a_i = a_{n-i}$, $n = 6$. The well-known method to compute the roots is to divide both sides by the middle term : $x^3$, and use the substitution: $y = x + \dfrac{1}{x}$ to reduce it to a lower degree.
A: If you put $y = \frac{1}{x}$ in the equation $f(x) = 0$, you shall get the same equation $f(y) = 0$ because of $a_i = a_{n-i}$. So you have the two following equations
$$f(x) = a_0 x^n + \dots a_n = 0 \\ f(y) = a_n y^n + \dots a_0 = 0.$$
Let we can write the solutions as $x_1 , \dots, x_n$. Then due to the second equation we can write then as $\frac{1}{x_1} , \dots , \frac{1}{x_n}$
So if $a$ is a root of such a equation $\frac{1}{a}$ must also be a root. 
A: The condition on the coefficients is equivalent to $x^n f(1/x) = f(x)$.
A: The first property that strikes me is this:
The product of its roots = $\frac{a_0}{a_n} = 1$, which means:


*

* The roots are of the form $x_i$ and $\frac{1}{x_i}$ 

* If $n$ is odd, $-1$ is a root of the equation.

* If $1$ is a root, then it has to be a double root, otherwise, the product of roots would be $-1$.


Will add more if I find any :)
