# How many roots are in case of polynomial $f(x).g(x)$

If two real polynomials $$f(x)$$ and $$g(x)$$ of degrees $$m(\geq 2)$$ and $$n(\geq 1)$$ respectively satisfy $$f(x^2+1)=f(x)g(x)$$ for every $$x \in R$$ then

(A) $$f$$ has exactly one real root $$x_0$$ such that $$f'(x_0)\neq 0$$

(B) $$f$$ has exactly one real root $$x_0$$ such that $$f'(x_0)= 0$$

(C) $$f$$ has $$m$$ distinct real roots

(D) $$f$$ has no real root

I tried finding value to this function and what I've figured out is that if $$f(x)$$ is of degree-2, then $$g(x)$$ should also be of degree-2 but I'm unable to find an equation of $$f(x)$$ where it solves $$\frac{f(x^{2}+1)}{f(x)}=g(x)$$

Note that if $$x_0$$ is a root of $$f(x)$$, then so is $$x_0^2+1$$. Also note that for all reals $$x$$, it holds that $$x^2+1>x$$. Thus, if $$x_0$$ is a real root of $$f(x)$$, then $$x_{n} = x_{n-1}^2+1$$, $$n\ge 1$$ gives infinitely many real roots of $$f(x)$$, which implies that $$f(x) = 0$$. This leads to a contradiction, and we conclude there is no real root of $$f(x)$$.