How do I find roots of a single-variate polynomials whose integers coefficients are symmetric wrt their respective powers Given a polynomial such as $X^4 + 4X^3 + 6X^2 + 4X + 1,$ where the coefficients are symmetrical, I know there's a trick to quickly find the zeros. Could someone please refresh my memory?

Such a polynomial with symmetric coefficients is called a palindromic polynomial. If $x_0$ is a zero of a palindromic polynomial than $\frac{1}{x_0}$ is a zero , too.
If we divide  the polynomial 
$$Ax^4+Bx^3+Cx^2+Bx+A=0$$
by $A$ we get 
$$x^4+ax^3+bx^2+ax+1=0 \tag{1}$$
$$(x^4+1)+a(x^3+x)+bx^2=0$$
$$x^2(x^2+\frac{1}{x^2})+ax^2(x+\frac{1}{x})+bx^2=0$$
$$x^2\left((x^2+\frac{1}{x^2})+a(x+\frac{1}{x})+b\right)=0$$
because $x \ne0$. For the same reason we can divide the equation by $x^2$  to get
$$(x^2+\frac{1}{x^2})+a(x+\frac{1}{x})+b=0  \tag{2}$$
We set 
$$y=x+\frac{1}{x} \tag{3}$$
and by squaring we get
$$y^2=x^2+\frac{1}{x^2}+2$$
If we substitute this in $(2)$ we get a quadratic equation of $y$ 
$$(y^2-2)+ay+b=0 \tag{4}$$
So $(1)$ can be solved by first solving the quadratic equation $(4)$ and then the quadratic equation $(3)$
In a similar  manner the equation 
$$x^4+ax^3+bx^2-ax+1=0 \tag{5}$$
can be solved:
$$(x^2+\frac{1}{x^2})+a(x-\frac{1}{x})+b=0$$
We set 
$$y=x-\frac{1}{x} \tag{6}$$
and now get
$$y^2=x^2+\frac{1}{x^2}-2$$
and so
$$(y^2+2)+ay+b=0 \tag{7}$$
So here we have to solve the quadratic equations $(7)$ and $(6)$ to get the solution of $(5)$.
An palindromic equation of degree $3$  can be solved by a different tick:
We transform
$$x^3+ax^2+ax+1=0 \tag{8}$$
to
$$(x^3+1)+a(x^2+x)=0$$
$$(x+1)\left((x^2-x+1)+ax\right)=0$$
So one solution is 
$$x=-1$$
This can also be seen  by plugging  $-1$ into $(8)$.
The two other solutions are the solutions of the quadratic equation
$$(x^2-x+1)+ax=0$$
Finally the equation 
$$x^3+ax^2-ax-1=0$$
can be transformed to 
$$(x^3-1)+a(x^2-x)=0$$
and therefore to
$$(x-1)\left((x^2+x+1)+ax\right)=0$$
An palindromic equation  of 6th or 9th degree again can be solved in two steps by solving first an equation of $y$ of degree $3$ or $4$ and then solve the equation
$$y=x+\frac{1}{x}$$
The palindromic equation of degree $5$  has the solution $x=-1$ and can be reduced to an equation of degree $$.
The polynomial equations up to degree $4$ can be solved by radicals, so the palindromic polynomial equations of degree  $5,6,8$ can be solved by radicals, too.
