Question

Solve the equation `x^4+4x^3+8=0`?

Obtain the resolvent cubics, by Descartes' method and by Ferrari's method, of the equation `x^4+4x^3+8=0`. Are the cubics the same? Further, Use either method to obtain the roots of this equation.

Answer 1

See below.

Explanation:

Consider the polynomial

`x^4+2ax^3+b=0`

and also the expansion

`(x^2+ax+xi)^2=x^4+2ax^3+(a^2+2xi)x^2+2axix+xi^2`

(here `xi` is a dummy parameter)

now substituting `x^4+2ax^3 = -b` we obtain

`(x^2+ax+xi)^2=(a^2+2xi)x^2+2axix+xi^2-b`

and now choosing properly `xi` to make `(a^2+2xi)x^2+2axix+xi^2-b` a square with

`2 xi^3- 2 b xi -a^2 b=0`

or

`xi^3-8xi-16=0`

///////////////////////////////////////////////////////////////////////////////////////////
Note that solving `(a^2+2xi)x^2+2axix+xi^2-b=0` or

`x=(-a xi pm sqrt[a^2 b + 2 b xi - 2 xi^3])/(a^2 + 2 xi)` if we choose
`xi_0` such that `a^2 b + 2 b xi_0 - 2 xi_0^3=0` then

`(a^2+2xi_0)x^2+2axi_0x+xi_0^2-b = (a^2+2xi_0)(x+a xi_0)^2`
////////////////////////////////////////////////////////////////////////////////////////////

Now solving for `xi`

`xi=xi_0=1/3 (216 - 24 sqrt[57])^(1/3) + (2 (9 + sqrt[57])^(1/3))/3^(2/3)` which is the only real root.

Now

`(a^2+2xi_0)x^2+2axi_0x+xi_0^2-b = (a^2+2xi_0)(x+ axi_0)^2`

and the former quartic remains

`(x^2+ax+xi_0)^2=(a^2+2xi_0)(x+ axi_0)^2` or

`(x^2+ax+xi_0)^2-(a^2+2xi_0)(x+ axi_0)^2=0` or

`(x^2+ax+xi_0+sqrt(a^2+2xi_0)(x+axi_0))(x^2+ax+xi_0-sqrt(a^2+2xi_0)(x+axi_0))=0`

resulting in the resolution of

`{(x^2+(a+sqrt(a^2+2xi_0))x+xi_0+axi_0sqrt(a^2+2xi_0)=0),(x^2+(a-sqrt(a^2+2xi_0))x+xi_0-axi_0sqrt(a^2+2xi_0)=0):}`

This is left as an exercise for the proficient reader.

NOTE: This is the so called Ferrari's method.