# Solving $a_1x^{4999} + a_2x^{4998} + a_3x^{4007}+…+a_{5000}x^{0}=n$

How can we solve the equation

$$a_1x^{4999} + a_2x^{4998} + a_3x^{4007}+...+a_{5000}x^{0}=n$$

if we know the values $a_1,a_2,a_3,...,a_{5000},n$? Are there any open source solutions?

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Tag description: DO NOT USE THIS TAG. The algebra tag is no longer being used. Please use the [algebra-precalculus] tag or the [abstract-algebra] tag instead. – Belgi Nov 22 '12 at 13:54
Out of curiosity, where did you get this problem? I assume you want a numerical solution, but it seems difficult to even evaluate this polynomial numerically. – littleO Nov 22 '12 at 13:55
of course it is from not-so-friendly geometric-like sequence and I solved it using WolframAlpha trial and verified it myself using binary search. And it turns out that solving the equation this way was wrong for this kind of problem. – thkang Nov 22 '12 at 14:16

## 1 Answer

Abel–Ruffini theorem tells us you can not solve (exactly) for the roots of this polynomial, since it is of degree $>4$

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This is true for the generic polynomial, but can be very wrong for specific polynomials. For instance, we know the exact roots of $1+x+x^2+\cdots+x^n=0$ for every $n$. – Andrea Mori Nov 22 '12 at 14:12
@AndreaMori ofcourse the Galois group of the polynomial can be solvable, but the OP was looking for an open sorce code and such a thing should give a correct outpit for every input. Numerical methods are the only real options, unless the case is very trivial. – Belgi Nov 22 '12 at 14:19
Absolutely. I guess that I was suggesting that although the situation looks bad in general, sometimes we do get lucky. – Andrea Mori Nov 22 '12 at 14:42
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