Ferrari's Method for Quartics

Date: 04/16/99 at 10:08:35
From: Stephane CATHELAIN
Subject: Ferrari's method

Could you briefly explain Ferrari's method for quartics or give me the 
location of a Web site where I can find more information on this 
method?

Thanks.

Date: 04/16/99 at 19:54:16
From: Doctor Anthony
Subject: Re: Ferrari's method

Ferrari (1540), a pupil of Cardano, was given a problem that could be 
stated in the following form:

  a+b+c = 10,     a/b = b/c,   ab = 6

which gives a = 6/b,  c = b^2/a = b^3/6

and then  6/b + b + b^3/6 = 10

    b^4 + 6b^2 - 60b + 36 = 0

a quartic without a third-degree term, which Ferrari rewrote

      (b^2 + 6)^2 =  6b^2 + 60b

and again after introducing a new variable z

    (b^2 + 6 + z)^2 = (2z + 6)b^2 + 60b + (12z + z^2)......(1)

For the righthand side of the equation to be a perfect square, the 
discriminant of the quadratic polynomial in b must equal zero. 

That is   4(2z+6)(z^2 + 12z) = 60^2

and this produces a cubic in z

     z^3 + 15z^2 + 36z - 450 = 0

By putting z = y-5 we get rid of the squared term and obtain

     y^3 - 39y - 380 = 0   which is the reducing cubic.

From here you proceed to the solution of a cubic. Having solved for y 
you get the values of z and then the righthand side of (1) is a 
perfect square. You now take the square root of both sides and have 
a quadratic in b.

There are other methods for dealing with quartics. The following is 
one such.

Example:  x^4 - 6x^3 + 15x^2 - 18x + 10 = 0

This is not a general method for a quartic but in the case of all real 
coefficients we know that it will have two pairs of conjugate roots. 
In looking for the quadratic factors it is reasonable with the given 
coefficients to assume that they must be of the form:

   (x^2 + ax + 5)(x^2 + bx + 2)  and multiplying out we get

   x^4 + x^3(a+b) + x^2(2 + 5 + ab) + x(2a+5b) + 10

   x^4 + x^3(a+b) + x^2(7+ab) + x(2a+5b) + 10    compare with

   x^4 + x^3(-6)  + x^2(15)   + x(-18)  + 10

comparing the coefficients

       a+b = -6     ab = 8     2a+5b = -18

if a = -4,  b = -2 these equations are all satisfied, so the quartic 
can be factorized into

    (x^2 -4x +5)(x^2 -2x + 2) = 0  

and then solving the equation  x^2 - 4x + 5 = 0 gives  x = 2 +-i
and solving       "            x^2 - 2x + 2 = 0 gives  x = 1 +-i

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/