<s>[INST] Let f(x) be a polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ Let f(x) be a  polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ .
And then there are some question related to this.
In the solution it is given that according to the given information, $f(x)$ is a polynomial of degree 4 with leading coefficient 3.
$f(x)=3(x-2)(x-3)(x+1)(x+6)+(x^2+1)$
Now what i think is that he got the coefficient 3 using the limits and the rest of the terms$[(x-2)(x-3)(x+1)(x+6)]$ from the respective $f(x)$ values from what was given and also the $(x^2+1)$ was added for the remainders, as they satisfy all the values that have been given to us in the question. I just wanted to know if my approach is right? or is there another elegant way to do it?.
 [/INST] the correct and precise reason for writing f(x)=3(x−2)(x−3)(x+1)(x+6)+(x^2+1) is as follows (I am assuming you have understood why f(x) will be a 4th deg polynomial and how its leading coeff shall be 3 from other answers):
consider g(x)= f(x)-(x^2+1)
clearly, deg(g(x))= 4
now, g(2)= f(2)-(2^2+1)= 0
similarly, g(3)=g(-1)=g(-6)= 0
hence, we have 4 roots of g(x) as 2, 3, -1, -6.
hence, by factor theorem g(x)= A(x−2)(x−3)(x+1)(x+6) where A is a constant.
or f(x)-(x^2+1)= A(x−2)(x−3)(x+1)(x+6)
or f(x)= A(x−2)(x−3)(x+1)(x+6)+x^2+1
but A=3 (already seen).
hence f(x)=3(x−2)(x−3)(x+1)(x+6)+(x^2+1)
</s>