How do you determine the number of complex roots of a polynomial of 3 degrees?

Suppose we are given `f(x) = ax^3+bx^2+cx+d` with `a > 0`

The slope of the tangent of the graph of `f(x)` at any `x` is given by the derivative:

`f'(x) = 3ax^2+2bx+c`

Using the quadratic formula, the cubic has turning points where `f'(x) = 0`, which is when

`x = (-2b+-sqrt(4b^2-12ac)) / (6a) = (-b+-sqrt(b^2-3ac))/(3a)`

If `b^2-3ac < 0` then `f'(x) = 0` only has Complex roots, so `f(x)` has no Real turning points and therefore `f(x) = 0` has exactly one Real root and two non-Real Complex roots.

If `b^2-3ac = 0` then `f'(x) = 0` has one repeated Real root, so `f(x)` has an inflection point, but is strictly monotonically increasing. So again `f(x) = 0` has exactly one Real root and two non-Real Complex roots, unless the inflection point is itself a root. If `f(-b/(3a)) = 0` then this is a threefold Real root and there are no non-Real Complex roots.

If `b^2-3ac > 0` then `f'(x) = 0` has two distinct Real roots:

`x_1 = (-b-sqrt(b^2-3ac))/(3a)`

`x_2 = (-b+sqrt(b^2-3ac))/(3a)`

We know that `f(x_2) < f(x_1)`, but there are still several possibilities:

(1) `f(x_1) < 0`, `f(x_2) < 0`

`f(x) = 0` has one Real root in `(x_2, oo)` and two non-Real Complex roots.

(2) `f(x_1) = 0`, `f(x_2) < 0`

`f(x) = 0` has a repeated Real root at `x = x_1`, and another Real root in `(x_2, oo)`.

(3) `f(x_1) > 0`, `f(x_2) < 0`

`f(x) = 0` has three distinct Real roots in `(-oo, x_1)`, `(x_1, x_2)` and `(x_2, oo)`.

(4) `f(x_1) > 0`, `f(x_2) = 0`

`f(x) = 0` has one Real root in `(-oo, x_1)` and a repeated Real root at `x = x_2`.

(5) `f(x_1) > 0`, `f(x_2) > 0`

`f(x) = 0` has one Real root in `(-oo, x_1)` and two non-Real Complex roots.