Depressed cubic roots

Question

According to Wikipedia, a depressed cubic has only one root and 2 imaginary roots. Is this true? Can a depressed cubic of the form $x^3+px+q=0$ have 2 or 3 real roots?

Edit: Here is a screenshot of the Wikipedia excerpt:

Answer 1

Yes, it can. Consider the polynomial having as roots $-3$, $1$ and $2$: since the sum of the roots is zero, the coefficient of $x^2$ is zero (Viète's formula): $$ (x+3)(x-1)(x-2)=0. $$

One can also have two roots (one is repeated, of course): $$ (x-4)(x+2)^2=0 $$

Or just one, if two roots are complex: $$ (x-2)(x+1-i)(x+1+i)=0. $$

Check doing the multiplications, if you don't trust Viète.

However, I can't find what you say in the linked Wikipedia page.

Answer 2

Note that, if a polynomial has a complex root, then its complex conjugate is also a root. Hence, complex roots always come in pairs. Therefore, any cubic will have either $1$ real and $2$ complex roots, or all real roots.

Now let's take a look at the discriminant of a monic, depressed cubic with roots $\alpha_1, \alpha_2, \alpha_3$:

$$\Delta = -4c^3 -27d^2 = (\alpha_1 - \alpha_2)^2(\alpha_1-\alpha_3)^2(\alpha_2-\alpha_3)^2$$

If $c$ is sufficiently large, then the discriminant of the depressed cubic is positive. This corresponds to $\alpha_1, \alpha_2, \alpha_3 \in \mathbb{R}$.

However, note that if we make $c$ is sufficiently small relative to $d$, then the discriminant of the depressed cubic is negative. This is only possible if two of the roots are complex.