Question

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

Solution

## $$f(x)=f(4-x)$$ ......replacing $$x$$ by $$x+2$$$$f(2+x)=f(2-x)$$ this show $$f(x)$$ is a symmetrical The graph is like shown in figure $$1$$$$f(x)$$ is a polynomial of degree $$8$$ so there are $$8$$ rootsIt 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 $$16$$.Mathematics

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