Question

How do you find all the real and complex roots of `3x^3+2x^2+x-4=0`?

Answer 1

Use Cardano's method to find Real zero:

`x_1 = 1/9(-2+root(3)(505+135sqrt(14))+root(3)(505-135sqrt(14)))`

and related Complex zeros.

Explanation:

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

This cubic has no rational zeros. It turns out to have one Real zero and a Complex conjugate pair of non-Real zeros.

To simplify it while avoiding some fraction arithmetic, first multiply it by `3^2*3^3 = 243` to find:

`243 f(x) = 729x^3+486x^2+243x-972`

`=(9x)^3+3(9x)^2(2)+x-4`

`=(9x+2)^3+15(9x+2)-1010`

Let `t=9x+2`

We want to solve:

`t^3+15t-1010 = 0`

This substitution to reduce a polynomial to one of simpler form is called a Tschirnhaus transformation.

To solve the simplified cubic, use Cardano's method, letting `t = u+v`

`u^3+v^3+3(uv+5)(u+v)-1010 = 0`

Substitute `v = -5/u` to eliminate the term in `(u+v)` and find:

`u^3-125/u^3-1010 = 0`

Multiply through by `u^3` to get:

`(u^3)^2-1010(u^3)-125=0`

Using the quadratic formula, we find:

`u^3 = (1010+-sqrt(1010^2-(4*1*-125)))/(2*1)`

`=(1010+-sqrt(1020100+500))/2`

`=(1010+-sqrt(1020600))/2`

`=505+-135sqrt(14)`

Since this is Real and the derivation was symmetric in `u` and `v`, we can use one of these values as `u^3` and the other as `v^3` to find the Real root of `t^3+15t-1010 = 0` is:

`t_1 = root(3)(505+135sqrt(14))+root(3)(505-135sqrt(14))`

and Complex roots:

`t_2 = omega root(3)(505+135sqrt(14))+omega^2 root(3)(505-135sqrt(14))`

`t_3 = omega^2 root(3)(505+135sqrt(14))+omega root(3)(505-135sqrt(14))`

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

Then `x = 1/9(t-2)`, so the zeros of the original cubic are:

`x_1 = 1/9(-2+root(3)(505+135sqrt(14))+root(3)(505-135sqrt(14)))`

`x_2 = 1/9(-2+omega root(3)(505+135sqrt(14))+omega^2 root(3)(505-135sqrt(14)))`

`x_3 = 1/9(-2+omega^2 root(3)(505+135sqrt(14))+omega root(3)(505-135sqrt(14)))`