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?

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.
