calculus - Showing a cubic equation has at most one real root or three real roots

Suppose we have $x^3 + \alpha x + \beta = 0$

Problem:

the cubic equation has one real root if $\alpha > 0$ and three real roots if $4 \alpha^3 + 27 \beta^2 < 0$.

Try:

Let $f(x) = x^3 + \alpha x + \beta$. One has that $f'(x) = 3 x^2 + \alpha$. We have critical points when

$$3x^2 + \alpha = 0 \iff x^2 = - \frac{ \alpha }{3}$$

Clearly, if $\alpha > 0$, then we dont have critical points and $f'(x) > 0$ thus always increasing. Since $\lim_{x \to \infty} f(x) = \infty$ and $\lim_{x \to -\infty} = - \infty$, then there must be some $c$ such that $f(c) = 0$, so we have one real root.

Now, if $\alpha < 0$, then we have critical points $x = \pm \sqrt{ -\alpha/3 }$. but, here I am stuck. Any help would be greatly appreaciated.