Polynomial of $6th$ degree $f(x)$ satisfies $f(x) = f(2-x)$ for all $x \in R$ and if $f(x) =0$ has $6$ roots

Question

A polynomial of $6th$ degree $f(x)$ satisfies $f(x) = f(2-x)$ for all $x \in R$ and if $f(x) =0$ has

$4$ distinct and two equal roots, then sum of roots of $f(x) = 0$ is

from $f(x) = f(2-x),$ replace $x\rightarrow (x+1)$

so $f(1+x) = f(1-x)$ (function $f(x)$ is symmetrical about $x=1$ line )

wan,t be able to go further, could some help me with this, thanks

Answer 1

Given $f(x)$ symmetric about $x=1$ with four distinct and two equal roots:

It can be said that the equal roots are equal to $1$

$$f(x)=\pm(x-1)^2(x-a)(x-b)(x-c)(x-d)$$

and due to symmetry the function has form,

$$f(x)=\pm(x-1)^2(x-(1-p))(x-(1+p))(x-(1-q))(x-(1+q))$$

and the roots are

$$1,1,1-p,1+p,1-q,1+q$$

where $p,q\ne0$.