Why can't we analytically solve a polynomial of degree >= 5?

Question

Is there an analytical way to solve equations of degree 5 or above without using iterating techniques or the factor theorem?

I know this is impossible, but why is it so? Can anyone give a simple explanation.


The way you usually solve algebraic problems and get the result in the form of:

x = (some function of the coefficients of the polynomial)

Answer 1

This is not a simple result - it was a hugely important problem for hundreds of years until the 1820s, when first Abel demonstrated it was impossible, and then Galois worked out the entire theory of solvable polynomials.

I can't think of a really simple way to explain it. Galois defined a way of mapping equations to a Galois group. You could just say as the size of the Galois group goes up its complexity goes up, and past 4 the complexity is such that there is no way to solve it.

I will be interested to see if someone else can come up with a clearer explanation.

Answer 2

to my knowledge there is no known analytical way to solve a polynomial of degree greater or equal to 5. It may be because the general Cubic equations Y=ax^3+bx^2+cx+d can be solved using elliptic functions which is an analytic function (or more precisely a meromorphic function) and quartics can be reduced to cubic functions and hence solved using elliptic functions. However, there is no analytic function that satisfies a polynomial of degree 5 and also it cannot be reduced to a quartic. In addition if you only require the solution to 0=ax^3+bx^2+cx+d, it is essentially connected with roots of elliptic functions and as these are analytic functions explicit formulae is availble for these.

Answer 3

You need a bit of Galois theory to follow the proof through, although the above answer contains a decent summary. Paolo Ruffini made the first proof (although it did contain a small gap which Abel filled in.) Do a search for 'Abel–Ruffini theorem' or 'Abel's Impossibility Theorem' for further details.

Answer 4

do you mean like x^5 + x^4 + x^3 + x^2 + x + 1, persay?

Well, I know that this graph would follow the behavior of an x^3 graph since it is an odd function. I do not know why we can not factor it by hand. We don't use these type of equations in school anyway, and if we do, it's more of a "common sense" type of question which isnt very detailed nor demanding.

Answer 5

For some Quintic equations you can solve them algebraically. However, it had been proved nearly 200 years ago that one cannot solve a general quintic or higher equation, algebraically.
Ie there is no general solution in the form such as in quadratic terms:

if ax^2+bx+c=0, then

x=(-b+/-sqrt(b^2-4ac))/2a

Answer 6

let s,t,u,v and w be the roots
of the equation
(x-s)(x-t)(x-u)(x-v)(x-w)=0
multiply that out and write in the form

ax^5+bx^4+cx^3+dx^2+ex+f=0

imagine the problem in trying to
work out the roots s,t,u,v and w
from the coefficients a,b,c,d,e
and f-it's a complete nightmare!

i hope that this helps