Question

How do you find all the real and complex roots of `f(x)= x^3-x+3`?

Answer 1

Use Cardano's method to find Real zero:

`x_1 = 1/6(root(3)(-324+12sqrt(717))+root(3)(-324-12sqrt(717)))`

and related Complex zeros.

Explanation:

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

Using the rational root theorem, we find that the only possible rational zeros of `f(x)` are: `+-1`, `+-3`. None of these are zeros, so `f(x)` has no rational zeros.

In fact it turns out to have one Real zero and two non-Real Complex zeros.

We can use Cardano's method to find all three as follows:

Let `x = u+v`. Then `f(x) = 0` becomes:

`u^3+v^3+(3uv-1)(u+v)+3 = 0`

Let `v=1/(3u)` to eliminate the term in `(u+v)` and get:

`u^3+1/(27u^3)+3 = 0`

Multiply through by `27u^3` to get:

`27(u^3)^2+81(u^3)+1 = 0`

Using the quadratic formula, we get:

`u^3=(-81+-sqrt(81^2-4*27*1))/(2*27)`

`=(-81+-sqrt(6561-108))/54`

`=(-81+-sqrt(6453))/54`

`=(-81+-3sqrt(717))/54`

`=(-324+-12sqrt(717))/216`

Since these roots are Real and the derivation was symmetric in `u` and `v`, we can use one of these roots for `u^3` and the other for `v^3` to find the Real zero:

`x_1 = 1/6(root(3)(-324+12sqrt(717))+root(3)(-324-12sqrt(717)))`

and Complex zeros:

`x_2 = 1/6(omega root(3)(-324+12sqrt(717))+omega^2 root(3)(-324-12sqrt(717)))`

`x_3 = 1/6(omega^2 root(3)(-324+12sqrt(717))+omega root(3)(-324-12sqrt(717)))`