If two real polynomials $f(x)$ and $g(x)$ of degrees $m(\geq 2)$ and $n(\geq 1)$ respectively, satisfy $$ f\left(x^{2}+1\right)=f(x) g(x) $$ for every $x \in \mathbb{R},$ then
(A) $f$ has exactly one real root $x_{0}$ such that $f^{\prime}\left(x_{0}\right) \neq 0$
(B) $f$ has exactly one real root $x_{0}$ such that $f^{\prime}\left(x_{0}\right)=0$
(C) $f$ has $m$ distinct real roots
(D) $f$ has no real root.
My approach
let f(x) has a real root $x_{0}$
$$ f\left(x_{0}^{2}+1\right)=f(x_{0})g(x_{0}) $$ $f\left(x_{0}^{2}+1\right)=0$
it's indicating $\mathrm{}, \quad\left(x_{0}^{2}+1\right)$ also a root of $\mathrm{f}(\mathrm{x})$
We can assume, $$ x=x^{2}+1; x^{2}-x+1=0; x=\frac{+1 \pm \sqrt{1-4}}{2}=\frac{+1 \pm \sqrt{3}}{2} $$ So,Not possible, $\left(x_{0}^{2}+1\right)$ also is a root ot $f(x)$ along With $x_{0}$.
If we go like this , let $x_{1}$ is a real root of $f(x)$
$f(x_{1}) = 0$ again $(x_{0}^{2}+1)$ will also be a root!
So, $f(x)$ will contain infinite root,but that is violating the condition of degree m. Hence $f(x)$ has no real roots.
You have shown that $x_0\neq 1+x_0^{2}$ but there is no guarantee that the third root is different from the first root $x_0$. Instead of this let $x_1=1+x_0^{2}$ and $x_{n+1}=1+x_n^{2}$ for all $n \geq 2$. Verify by induction that $x_n \geq n$ for all $n \geq 1$. This shows that $x_n \to \infty$ so there are infintely many roots.