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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.

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Supriyo Banerjee
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    @Mason sorry, it has been edited – Supriyo Banerjee Apr 24 '20 at 17:32
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    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. – Paul Sinclair Apr 24 '20 at 22:55
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    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 $x_0^2 + 1 = x_0$... – Paul Sinclair Apr 24 '20 at 23:07

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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.

Paul Sinclair
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