Question

What are all the zeroes of `f(x)=5x^3+8x^2-4x+3`?

Answer 1

The Real root is:

`x_1 = -8/15+root(3)((-4489+45sqrt(6185))/6750)+root(3)((-4489-45sqrt(6185))/6750)`

`~~ -2.11299545715260`

Complex roots as shown below.

Explanation:

`f(x)=5x^3+8x^2-4x+3`

This is of the form `ax^3+bx^2+cx+d` with `a=5`, `b=8`, `c=-4` and `d=3`

Its discriminant `Delta` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

`=1024+1280-6144-6075-8640=-18555`

Since `Delta < 0` this cubic has one Real zero and two Complex conjugate ones.

So we can use a Tschirnhaus transformation to eliminate the `x^2` term, then use Cardano's method.

Let `t = x+8/15`

Then:

`5t^3 = 5(x+8/15)^3 = 5x^3+8x^2+64/15x+512/675`

`-124/15t = -124/15(x+8/15) = -124/15x-992/225`

Hence:

`5t^3-124/15t+4489/675 = 5x^3+8x^2-4x+3`

Multiplying through by `675`, we want to solve:

`3375t^3-5580t+4489 = 0`

Let `t=u+v`

`3375u^3+3375v^3+(10125uv-5580)(u+v)+4489=0`

To eliminate the `(u+v)` term add the constraint:

`v = 5580/(10125u) = 124/(225u)`

Then we have:

`0 = 3375u^3+3375(124/(225u))^3+4489`

`=3375u^3+3375(1906624/(11390625u^3))+4489`

`=3375u^3+1906624/(3375u^3)+4489`

Multiply through by `3375u^3` to get:

`11390625(u^3)^2+15150375(u^3)+1906624 = 0`

Then by the quadratic formula:

`u^3 = (-15150375+-sqrt(15150375^2-(4*11390625*1906624)))/(2*11390625)`

`= (-15150375+-sqrt(142663306640625))/22781250`

`= (-15150375+-151875sqrt(6185))/22781250`

`= (-4489+-45sqrt(6185))/6750`

Since the derivation is symmetric in `u` and `v`, we can deduce:

`t = root(3)((-4489+45sqrt(6185))/6750)+root(3)((-4489-45sqrt(6185))/6750)`

`x_1 = -8/15+root(3)((-4489+45sqrt(6185))/6750)+root(3)((-4489-45sqrt(6185))/6750)`

The Complex roots are:

`x_2 = -8/15+omega root(3)((-4489+45sqrt(6185))/6750)+omega^2 root(3)((-4489-45sqrt(6185))/6750)`

`x_3 = -8/15+omega^2 root(3)((-4489+45sqrt(6185))/6750)+omega root(3)((-4489-45sqrt(6185))/6750)`

where `omega = -1/2+sqrt(3)/2i`