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This question is an exact duplicate of:

If $f(x)$ Polynomial with real coefficient and $f(0)=1$, $f(2)+f(3)=125$ and $f(x)*f(2x^2)=f(2x^3+x)$

then what is the value of $f(5)$?

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What I tried $$f(0)=1$$ put x=1 $$f(1)*f(2)=f(3)$$ put x=2 $$f(2)*f(8)=f(18)$$

from this approach, I cant find f(5). Please Suggest a method to solve.

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marked as duplicate by Dietrich Burde, mrtaurho, Davide Giraudo, drhab, Adrian Keister Apr 26 at 14:48

This question was marked as an exact duplicate of an existing question.

  • $\begingroup$ I edit the question. That's all i know about the question $\endgroup$ – Pankaj Solanki Apr 25 at 13:27
  • $\begingroup$ Trying $x=1$ and $x=2$ is probably not enough. You should try more. $\endgroup$ – Dietrich Burde Apr 25 at 13:35
  • $\begingroup$ does * represent multipication operator? $\endgroup$ – Marvel Maharrnab Apr 25 at 15:47
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Here's a somewhat unfulfilling solution.

According to Solve $f(x)f(2x^2) = f(2x^3+x)$, the polynomial $f(x)$ is of the form $f(x)=(x^2+1)^n$ for some $n$.

Using $f(2)+f(3)=125$ we obtain $n=2$, so $f(x)=(x^2+1)^2$. Plugging in $x=5$ we conclude $$f(5)=(5^2+1)^2=26^2=676.$$

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  • $\begingroup$ I check the math.stackexchange.com/questions/2600903/solve-fxf2x2-f2x3x page. How to get that $f(x)=(x^2+1)^n$ is the form of $f(x)$. Would you please explain it to me. You can give me Hint also. $\endgroup$ – Pankaj Solanki Apr 25 at 16:21
  • $\begingroup$ @PankajSolanki The accepted solution is quite complete and doesn't use any advanced facts. If you have difficulty understanding a specific point I would be glad to help you out. $\endgroup$ – Luiz Cordeiro Apr 25 at 17:26
  • $\begingroup$ I got my answer. Thanks for Help. $\endgroup$ – Pankaj Solanki Apr 26 at 11:16

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