<s>[INST] Method to find the unique substitution which interchanges the original polynomials in this problem If $f$ is an $R$ to $R$ function such that $f(x^2+x+3)+2f(x^2-3x+5)=6x^2-10x+17$ find f(x)

One way would be to assume its a polynomial equation and solve for it . Otherwise              : Notice that if we replace $x$ by $1-x$ then $f(x^2-3x+5)+2f(x^2+x+3)=6x^2-2x+13$
Now we just need to solve the system of equations:
$\left\{\begin{matrix}
f(x^2+x+3)+2f(x^2-3x+5)=6x^2-10x+17\\ 
f(x^2-3x+5)+2f(x^2+x+3)=6x^2-2x+13
\end{matrix}\right.$
Solving that system of equations gives $\left\{\begin{matrix}
f(x^2+x+3)=2x^2+2x+3\\ 
f(x^2-3x+5)=2x^2-6x+7
\end{matrix}\right.$
This gives $f(x)=2x-3$.


*

*My question is how to deduce /calculate and arrive at that there is a way (here its $x \rightarrow 1-x$ ) such that two polynomial inside the function as input can be interchanged into each other via unique substitution ? And what are these pair of polynomials called as ? They seem to be very special in the sense everything about their nature is deducable from its other one .

 [/INST] You are looking for a substitution $x \mapsto g(x)$ such that $\,g^2(x) + g(x) + 3 = x^2 - 3x + 5\,$ $\,\iff g(x)\left(g(x) + 1\right) = x^2-3x+2=(x-2)(x-1)=(1-x)(2-x)\,$. The factors on both sides differ by $\,1\,$, so the natural substitutions to try are $\,g(x)=x-2\,$ or $\,g(x) = 1-x\,$. They both work, but only $\,g(x) = 1-x\,$ also satisfies $\,g^2(x)- 3 g(x) + 5 = x^2+x+3\,$ which is required for the reverse substitution to work.
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