Newton's method

Galois showed that there is no Algebraic (finite) formula for the solution of a general fifth or higher degree polynomial equation. The formulae for the first, second, third, and fourth degrees had been know, already.

Linear (first degree) a x + b = 0 has the root x = - b / a.

Quadratic (second degree) a x^2 + b x + c = 0 has the pair of roots given by either x = (- b +/- sqrt(b^2 - 4 a c)) / (2 a) or x = (2 c) / (- b -/+ sqrt(b^2 - 4 a c)). Which should we employ? It is a matter of round off errors. If b > 0 use the minus signs. If b < 0 use the plus signs. Otherwise (that is, if b = 0), both reduce to +/- sqrt(- c / a).

For the cubic and quartic, we will take a0 = 1 (monic) and a1 = 0 (average value of the roots is zero).

Cubic (third degree) x^3 + a x + b = 0 has the three roots (courtesy of Macsyma):

x = (- sqrt(3) i / 2 - 1 / 2) u - (sqrt(3) i / 2 - 1 / 2) a / (3 u)

x = (sqrt(3) i / 2 - 1 / 2) u - (- sqrt(3) i / 2 - 1 / 2) a / (3 u)

x = u - a / ( 3 u)

where

u = cuberoot(sqrt(27 b^2 + 4 a^3) / (6 sqrt(3)) - b / 2)

Quartic (fourth degree) x^4 + a x^2 + b x + c = 0 has the four roots (again, courtesy of Masyma):

Sorry, it just is too complicated. You will have to ask Macsyma for yourself. :-:)

where

u = cuberoot(sqrt(- 256 c^3 + 128 a^2 c^2 - (144 a b^2 + 16 a^4) c + 27 b^4 + a^3 b^2) / (6 sqrt(3) + (- 72 a c + 27 b^2 + 2 a^3) / 54)

The Newton method takes the Taylor's series

y = yo + y'o (x - xo) + y''o (x - xo)^2 / 2 + y'''o (x - xo)^3 / 6 + y''''o (x - xo)^4 / 24 + ....

Sets y = 0 and solves for the next iterative value of x. The higher the degree the faster it converges; but becomes rapidly harder to solve. In practice, the complexity of finding the derivatives must be balanced against the completely of solving the equation. The Newton methods for quartic or cubic are conceivable; but not practical.