# 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 $f(5)$? [duplicate]

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.

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

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

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

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 $$f(5)=(5^2+1)^2=26^2=676.$$
• 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. – Pankaj Solanki Apr 25 at 16:21