Conditions for distinct real roots of cubic polynomials.

Given a cubic polynomial with real coefficients of the form $f(x) = Ax^3 + Bx^2 + Cx + D$ $(A \neq 0)$ I am trying to determine what the necessary conditions of the coefficients are so that $f(x)$ has exactly three distinct real roots. I am wondering if there is a way to change variables to simplify this problem and am looking for some clever ideas on this matter or on other ways to obtain these conditions.

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Suppose that (including multiplicity) the roots of $$f(x) = A x^3 + B x^2 + C x + D,$$ $A \neq 0$ are $r_1, r_2, r_3$. Then, inspection shows that the quantity
$$D(f) := A^4 (r_3 – r_2)^2 (r_1 – r_3)^2 (r_2 – r_1)^2,$$
called the (polynomial) discriminant of $f$, vanishes iff $f$ has a repeated root. (The coefficient $A^4$ is unnecessary for the expression to enjoy this property, but among other things, its inclusion makes the below formula nicer.) On the other hand, with some work (say, by expanding and using Newton’s Identities and Vieta’s Formulas) we can write $D(f)$ as a homogeneous quartic expression in the coefficients $A, B, C, D$:
$$D(f) = -27 A^2 D^2 + 18 ABCD – 4 A C^3 – 4 B^3 D + B^2 C^2.$$

It turns out that $D$ gives us the finer information we want, too: $f$ has three distinct, real roots iff $D(f) > 0$ and one real root and two conjugate, nonreal roots iff $D(f) < 0$.

It’s apparent that one can generalize the notion of discriminant to polynomials $p$ of any degree $> 1$, producing an expression homogeneous of degree $2(\deg p – 1)$ in the polynomial coefficients. In each case, up to a constant that depends on the degree and the leading coefficient of $f$, $D(f)$ is equal to the resultant $R(f, f’)$ of $f$ and its derivative $f'(x) = 3 A x^2 + 2 B x + C$.

By making a suitable affine change of variables $x \rightsquigarrow y$, by the way, one can transform the given cubic to the form
$$\tilde{f}(y) = y^3 + P y + Q$$ (which does not change the multiplicity of roots), and for a cubic polynomial in this form the discriminant has the simple and well-known form
$$-4 P^3 – 27 Q^2.$$

By solving $f'(x)=0$, you can find out the two turning points of the cubic. Suppose the solutions are $x_1$ and $x_2$. In fact, we have
$$
x_{1,2} = \frac{-B \pm \sqrt{B^2-3AC}}{3A}
$$

If $B^2-3AC <0$, then $x_1$ and $x_2$ are imaginary, so there are no turning points, and the cubic has only one real root.

If $x_1$ and $x_2$ are real, the cubic has three distinct roots iff $f(x_1)$ and $f(x_2)$ are non-zero and have opposite sign.

There are numerous corner cases to worry about, but that’s the basic idea.

Edit: Following up on the comment above, the Wikipedia page says that the nature of the roots can be determined by examining the discriminant:
$$
\Delta = 18ABCD – 4B^3D + B^2C^2 -4AC^3 – 27A^2D^2
$$
The cubic has three distinct real roots iff $\Delta > 0$. This is a nice tidy criterion, but my discussion above may still be useful because it provides some geometric insight.

For simplicity assume $A>0$ (you can easily get analogous conditions for $A<0)$. Then the first turning point will be a maximum and the second one will be a minimum. Then the 3 distinct roots:

(i) the least root should be earlier than the point where maximum is attained

(ii) the middle root should be between the maximum and the minimum

(iii) the largest root would be bigger than the minimum.

This translates to maximum should be attained at a negative number, minimum at a positive number.
You can now express the above statement into a condition on the coefficients of $f’x$.