Question

If $f\left(x\right)$ be a polynomial of degree $8$ such that $f\left(x\right)=f(4-x)$ $\forall xϵR$, $f\left(x\right)$ has $6$ distinct real and equal roots then sum of roots of $f\left(x\right)=0$ is $A$. Then the number $A$ is:

Answer 1

$f\left(x\right)=f(4-x)$ ......replacing $x$ by $x+2$
$f(2+x)=f(2-x)$ this show $f\left(x\right)$ is a symmetrical
The graph is like shown in figure $1$
$f\left(x\right)$ is a polynomial of degree $8$ so there are $8$ roots
It has two equal root which is $2$ , $2$ and six distinct root in form of $2+a$ , $2-a$, $2+b$, $2-b$, $2+c$, $2-c$
Now sum of roots are $2+2+2+a+2-a+2+b+2-b+2+c+2-c$
Sum of roots is $16$.