$f(x^2 + 1) = f(x)g(x)$ $\forall$ x $\in\mathbb{R}$ $\Rightarrow$ no. of roots of $f(x)=$? If two real polynomials $f(x)$ and $g(x)$ of degrees $m$ $(\ge2)$ and $n$ $(\ge1)$ respectively, satisfy $f(x^2 + 1) = f(x)g(x)$; for every $x \in \mathbb{R}$ , then what can be said about the root(s), if any, of $f(x)$?

Clearly, if $x$ is a root then so is $x^2+1$.
Thus if $x_0$ is the largest real root, then we need $x_0\ge x_0^2+1$, that is $x_0^2-x_0+1\le0$. But $x_0^2-x_0+1=(x_0-\frac12)^2+\frac34>0$
