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?
 A: You can write $y=x+\frac{1}{x}$ and cut the degree in half.  In this case (having checked that $x$ cannot be $0$) $x^2+4x+6+\frac{4}{x}+\frac{1}{x^2}=y^2+4y+4$
A: I'm not sure if this is what you are thinking of, but if $x$ is a root, and $y=1/x$, then by plugging in $1/y$ and multiplying the result by $y^n$ (where $n$ is the degree of the polynomial), we see that $y$ is also a root of the polynomial.  This has two consequences.  First, every root, except possibly for $X=\pm 1$ comes in pairs, and so if the polynomial is of odd degree, it must have $\pm 1$ as a root.  Second, if it is of even degree, then for every root $x$, the polynomial is divisible by $(X-x)(X-1/x)=X^2-X(x+1/x)+1$.  This indicates that we should be able to rewrite the polynomial in terms of $X+1/X$.  We can do this as follows.
Let the polynomial be $P(X)=\sum_{i=0}^{n} a_i x^i$, where $a_i=a_{n-i}$  If $P(X)$ is of odd degree, we know that $1$ must be a root.  Divide out by $X-1$ and $X+1$ until $\pm 1$ are no longer roots, and you will get an even degree polynomial with symmetric coefficients, so from here on we can assume that $n$ is even.
Divide $P(X)$ by $X^{n/2}$ to get $P(X)/X^{n/2}=a_n(X^{n/2}+1/X^{n/2})+a_{n-1}(X^{n/2-1}+1/X^{n/2-1})+\ldots$.  We can write $X^k+X^{-k}$ as a polynomial in $(X+1/X)$ with symmetric coefficients, and subtracting off this chunk and repeating with the lower degree terms allows us to eventually rewrite $P(X)/X^{n/2}=Q(Z)$ where $Z=X+1/X$.  Thus, we have reduced the problem to one of half the degree.
Note that we can simplify the division step with a little calculus by noting that $1$ is a root of $P(X)$ of multiplicity $k$ if $P(1)=P'(1)=P''(1)=\ldots P^{(k-1)}(1)=0$, but $P^{(k)}(1)\neq 0$, and similarly for $P(-1)$.  
Also, it is worth pointing out that this is very similar to how, when you have a real polynomial, all the complex roots have to come in pairs with their conjugates, and so once you get rid of all the real roots, you can write the polynomial as a product of quadratic factors.  However, the fact that we can rewrite our polynomial here in terms of $X+1/X$ has no real analogue.
A: 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 $$.
The polynomial equations up to degree $4$ can be solved by radicals, so the palindromic polynomial equations of degree  $5,6,8$ can be solved by radicals, too.
A: If you use the Rational Root Theorem, you find that the only possible rational roots are $\pm 1$.  You can check if either of those work and if so, divide by the corresponding factor.  Then you get a simpler polynomial, and in your specific case you would be able to use the Rational Root Theorem again to get a quadratic.
EDIT: The Rational Root Theorem says that if you have a polynomial with integer coefficients, and that polynomial has a rational root $\frac{a}{b}$ where $\frac{a}{b}$ is in lowest terms, then $a$ divides the constant term and $b$ divides the leading coefficient.
A: Hint:  This particular polynomial is very nice, and factors as $(X+1)^4$. 
Take a look at Pascal's Triangle and the Binomial Theorem for more details.
Added:  Overly complicated formula
The particular quartic you asked about had a nice solution, but lets find all the roots of the more general  $$ax^{4}+bx^{3}+cx^{2}+bx+a.$$ Since $0$ is not a root, we are equivalently finding the zeros of
$$ax^{2}+bx^{1}+c+bx^{-1}+ax^{-2}.$$Let $z=x+\frac{1}{x}$ (as suggested by Aryabhatta) Then $z^{2}=x^{2}+2+x^{-2}$ so that $$ax^{2}+bx^{1}+c+bx^{-1}+ax^{-2}=az^{2}+bz+\left(c-2a\right).$$ The roots of this are given by the quadratic formula: $$\frac{-b+\sqrt{b^{2}-4a\left(c-2a\right)}}{2a},\ \frac{-b-\sqrt{b^{2}-4a\left(c-2a\right)}}{2a}.$$ Now, we then have $$x+\frac{1}{x}=\frac{-b\pm\sqrt{b^{2}-4a\left(c-2a\right)}}{2a}$$
and hence we have the two quadratics $$x^{2}+\frac{b+\sqrt{b^{2}-4a\left(c-2a\right)}}{2a}x+1=0,$$ $$x^{2}+\frac{b-\sqrt{b^{2}-4a\left(c-2a\right)}}{2a}x+1=0.$$ This then gives the four roots:$$\frac{-b+\sqrt{b^{2}-4a\left(c-2a\right)}}{4a}\pm\sqrt{\frac{1}{4}\left(\frac{b-\sqrt{b^{2}-4a\left(c-2a\right)}}{2a}\right)^2-1}$$
$$\frac{-b-\sqrt{b^{2}-4a\left(c-2a\right)}}{4a}\pm\sqrt{\frac{1}{4}\left(\frac{b+\sqrt{b^{2}-4a\left(c-2a\right)}}{2a}\right)^2-1}.$$
If we plug in $a=1$, $b=4$, $c=6$, we find that all four of these are exactly $1$, so our particular case does work out.
A: One possibility: Divide by $X^2$ and write it as a polynomial in $Z = X + 1/X$.
