Question

What are all the zeroes of `f(x) = 2x^3 - 5x^2 + 3x - 1`?

Answer 1

The Real root of `f(x) = 0` is:

`x_1 = 1/6 (5 + root(3)(44+3sqrt(177))+root(3)(44-3sqrt(177)))`

and Complex roots as below...

Explanation:

Given:

`f(x) = 2x^3-5x^2+3x-1`

First substitute `t = x-5/6`

Then:

`2t^3 = 2(x-5/6)^3 = 2(x^3-5/2x^2+25/12x-125/216)`

`=2x^3-5x^2+25/6x-125/108`

`color(white)()`

`-7/6t = -7/6x+35/36 = -7/6x+105/108`

`color(white)()`

`2t^3-7/6t = 2x^3-5x^2+3x-5/27`

So:

`2t^3-7/6t-22/27 = 2x^3-5x^2+3x-1 = f(x)`

Multiply through by `54` to get integer coefficients:

`108t^3-63t-44 = 54 f(x)`

So we want to solve:

`108t^3-63t-44 = 0`

Use Cardano's method, letting `t = u + v`

`0 = 108(u+v)^3-63(u+v)-44`

`= 108u^3+108v^3+(324uv-63)(u+v)-44`

`= 108u^3+108v^3+9(36uv-7)(u+v)-44`

Next make the coefficient of the `(u+v)` term zero by adding the constraint: `36uv-7 = 0`, that is `v = 7/(36u)`

`= 108u^3+108(7/(36u))^3-44`

`= 108u^3+343/(432u^3)-44`

Multiply through by `432 u^3` to get a quadratic in `u^3`:

`46656(u^3)^2-19008(u^3)+343 = 0`

Then using the quadratic formula:

`u^3 = (19008+-sqrt(19008^2-4*46656*343))/(2*46656)`

`= (19008+-sqrt(361304064-64012032))/93312`

`= (19008+-sqrt(297292032))/93312`

`= (19008+-1296 sqrt(177))/93312`

`= (44+-3sqrt(177))/216`

Since the derivation was symmetric in `u` and `v`, one of these roots is suitable for `u^3` and the other for `v^3`.

Hence the Real root is:

`t = root(3)((44+3sqrt(177))/216)+root(3)((44-3sqrt(177))/216)`

`= 1/6 (root(3)(44+3sqrt(177))+root(3)(44-3sqrt(177)))`

and hence:

`x_1 = 5/6 + t = 1/6 (5 + root(3)(44+3sqrt(177))+root(3)(44-3sqrt(177)))`

The Complex roots are:

`x_2 = 1/6 (5 + omega root(3)(44+3sqrt(177))+omega^2 root(3)(44-3sqrt(177)))`

`x_3 = 1/6 (5 + omega^2 root(3)(44+3sqrt(177))+omega root(3)(44-3sqrt(177)))`

where `omega = -1/2 +sqrt(3)/2 i` is the primitive Complex cube root of `1`.