# Which option is correct

If two real polynomials $$f(x)$$ and $$g(x)$$ of degrees m ($$\gt$$ $$1$$) and n ($$\gt$$ $$0$$) respectively,satisfy

$$f(x^2 + 1)$$ $$=$$ $$f(x)g(x)$$

for every $$x$$ $$\in$$ $$\mathbb R$$,then

Which one is correct

1. $$f$$ has exactly one real root $$x_0$$ such that $$f'(x_0)\ne 0$$.

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

3. $$f$$ has m distinct real roots.

4. $$f$$ has no real root.

I tried using some examples but failed.

• @Mason sorry, it has been edited Apr 24, 2020 at 17:32
• This is only possible when $m = n$. The left hand side is a polynomial of degree $2m$ while the right hand side is a polynomial of degree $m + n$. Polynomials of differering degrees cannot be equal for all real numbers. Apr 24, 2020 at 22:55
• Note that if $f(x_0) = 0$ and $f'(x_0) = 0$, then $f(x_0^2 + 1) = 0$ and $f'(x_0^2 + 1) = 0$, so for (2) to be true, you'd need $x_0^2 + 1 = x_0$... Apr 24, 2020 at 23:07

Note that if $$x$$ is a root of $$f$$, then so is $$x^2 + 1$$. But $$|x^2 + 1| > |x|$$ for all real numbers $$x$$. So every real root of $$f$$ requires the existence of another real root of higher magnitude.