What is the meaning of "due to the symmetry of the coefficients, if $x=r$ is a zero of $x^4+x^3+x^2+x+1$ then $x=\frac1r$ is also a zero" I was studying this answer about factoring $x^4+x^3+x^2+x+1$:
https://socratic.org/questions/how-do-you-factor-x-4-x-3-x-2-x-1
The author says: "A cleaner algebraic approach is to notice that due to the symmetry of the coefficients, if $x=r$ is a zero of $x^4+x^3+x^2+x+1$ then $x= {1\over r}$ is also a zero"
And eventually he writes  $x^4+x^3+x^2+x+1=(x^2+ax+1)(x^2+bx+1)$
Question $1$: What is the meaning of symmetry of the coefficients?
Question $2$: Can we do the same approach for $x^4-x^3+x^2-x+1$? ( I ask because it is relevant to my other question: Problem with factoring $x^4-x^3+x^2-x+1$)
 A: The list of coefficients of$$x^4+x^3+x^2+x+1$$is $(1,1,1,1,1)$, which is symmetric (if you reverse it, you will get the same list). In other words, it is a list of the type $(a,b,c,b,a)$. And if $r(\ne0)$ is a root of$$ax^4+bx^3+cx^2+bx+a,\tag1$$then$$ar^4+br^3+cr^2+br+a=0,$$and therefore$$a+\frac br+\frac c{r^2}+\frac b{r^3}+\frac a{r^4}=0$$too; in other words, $\frac1r$ is also a root of $(1)$. So, unless one of the roots is $\pm1$ (which are the only numbers equal to their own inverses), $(1)$ can be written as\begin{multline}a(x-r)\left(x-\frac1r\right)(x-r')\left(x-\frac1{r'}\right)=\\=a\left(x^2-\left(r+\frac1r\right)x+1\right)\left(x^2-\left(r'+\frac1{r'}\right)x+1\right).\end{multline}
In particular, $x^4-x^3+x^2-x+1$ can be written as$$(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1.$$In order to find $a$ and $b$, solve the system$$\left\{\begin{array}{l}a+b=-1\\ab+2=1.\end{array}\right.$$
A: To answer the original question, the thinking process comes as follows:
(1) If $r$ is a solution to $x^4-x^3+x^2-x+1=0$, then $r^4-r^3+r^2-r+1=0$.
(2) Divide both sides by $r^4$ you get $({1\over r})^4-({1\over r})^3+({1\over r})^2-({1\over r})+1=0$. Therefore $1\over r$ is also a solution.
(3) Hence if $(x-r)$ is a factor of the polynomial then $(x-{1\over r})$ is also a factor.
(4) Therefore the equation can be written as $(x-r)(x-{1\over r})(x-s)(x-{1\over s})$
(5) Therefore it can be written as $(x+ax+1)(x+bx+1)$
