Let $f(x)$ be a polynomial satisfying $\lim_{x\to \infty} \frac {x^4 f(x)}{x^8+1} =3$, $f(2)=5$, $f(3)=10$,$f(-1)=2$,$f(-6)=37$. Find $f(0)$ It’s clear that $f(x)$ is a 4th degree polynomial.
If $f(x)=ax^4+bx^3+cx^2+dx+e$, then $a=3$
From the rest of the given data, I can form four linear equations, which should give me the value of $a,b,c,d$, but that’s far too tedious and time consuming, and I don’t think the question is meant to be solved that way. Is there any alternative method?

It is clear that $f(x)$ is of deegree 4 and
$$\begin{align} f(2) &= 1 + 2^2
\\ f(-1) &= 1 + (-1)^2
\\ f(3) &= 1+ 3 ^ 2
\\ f(-6) &= 1+ (-6)^2 \end{align} $$
so $f(x) - (1+x^2)$ has roots $2,3,-1,-6$
$$ \Rightarrow f(x) - (1+ x^2) = A(x+1)(x-2)(x+6)(x-3) $$
now you can continue
