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


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 $1$


and due to symmetry the function has form,


and the roots are


where $p,q\ne0$.

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