$f(x^2 + 1) = f(x)g(x)$ $\forall$ 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 $f(x)$?

Hint: If $x$ is a root of $f$, then $f(x) = 0$. That means that $f(x^2 +1) = f(x)g(x) = 0$. So that means that $x^2 + 1$ is a solution. But $x < x^2 + 1$. You can see by noting that $x = x^2 +1$ doesn't have any roots and since $1^2 + 1 > 1$ you have $x < x^2 +1$ for all real numbers $x$.
So you must now have two roots, so ... 
