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In layman terms, why is it not possible to algebraically solve polynomials of degree five and higher?

Ecir Hana
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    I suspect there is no good answer in "layman terms", Mathematicians (Abel and Galois) had to invent some deep new theories to prove it. You need about a semester of college level abstract algebra to see a proof. See http://math.stackexchange.com/questions/116814/what-does-insolvability-of-the-quintic-mean-exactly – Ethan Bolker Mar 30 '17 at 18:48
  • https://www.youtube.com/watch?v=cxNq-hQwvn0 and https://www.youtube.com/watch?v=pdYe4BKcm74 – Count Iblis Mar 30 '17 at 22:41

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One author phrased it thusly (more or less): In a polynomial, every root is an irrational function of the coefficients, but every coefficient is only a rational function of the roots. This dichotomy sets up a sort of 'internal tension' between the roots and the coefficients, which increases with the degree of the polynomial. By degree 5 the 'tension' has exceeded the representational capacity of nested root-extractions.

PMar
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