When are the roots of a polynomial of degree 3 aligned?

Question

Let $P \in \mathbb{C}[X]$ be a polynomial of degree 3. On what condition on the coefficients of $P$ are the three roots of $P$ aligned ?

To make things easier, we may assume that $P$ can be written as $P = X^3 + aX^2 + bX + c$.

Let $z_1, z_2, z_3$ be the roots of $P$.

I tried to say that the condition was $\Im((a - c) \overline{(b - c)}) = 0$, but I could not manage to get anywhere. Trying to write $z_3 = k z_1 + (1 - k) z_2$ ($k \in [0, 1]$) did not help me either.

Thank you for your help.

Answer 1

Hint:

The derivative of

$$P(x)=x^3 + ax^2 + bx + c \tag{1}$$

is

$P'(x)= 3x^2+2ax+b \tag{2}$

Its two roots are

$$z'_k=\frac13(-a\pm d) \ \ \text{where} \ d:=\sqrt{a^2-3b}\tag{3}$$

(the square root that has to be taken in the complex sense).

These roots are known to be inside triangle $z_1z_2z_3$ (Gauss-Lucas' theorem).

The alignment of $z_1,z_2,z_3$ is therefore necessarily along the line $L$ defined by $[z'_1,z'_2]$ which can be described as the following set of points:

$$z=\lambda z'_1+(1-\lambda) z'_2= - \ \frac{a}{3}+\frac13(2 \lambda d -1)\tag{4}$$

where $\lambda \in \mathbb{R}$.

Now plug relationship (4) into (1).

Remark: a supplementary condition is the fact that two of the roots must lie ouside open segment $(z'_1,z'_2)$, implying that $0<\lambda<1$ can happen for at most one root.

Answer 2

(Not an answer, but too long for a comment. Prompted by comments under this other question.)

For an $\,n^{th}\,$ degree polynomial with $\,n \ge 3\,$, the question whether its roots lie on a line in the complex plane relates to the question of determining whether the roots of an associated polynomial are all real.

If the roots of $\,n^{th}\,$ degree $\,P(z)\,$ are collinear, then the roots of all its derivatives must also lie on the same line by the Gauss–Lucas theorem.

Taking some liberty in assuming that all roots are distinct, it follows that the line on which all roots lie is the line through the two roots $\,a,b\,$ of the quadratic $\,P^{(n-2)}(z)=0\,$. That line can be mapped onto the real axis with a translation and/or rotation, so a necessary condition for the roots of $\,P(z)\,$ to be collinear is for all roots of $\,Q(z) = P(\omega z) - \nu\,$ to be real, for the respective $\,\omega, \nu \in \mathbb C\,$, with $\,|\omega|=1\,$, both of which can be explicitly determined from $\,a,b\,$.

In the case $\,n=3\,$ of a cubic, the quadratic in question is $\,P'(z)\,$, and that case is covered nicely in Jean Marie's answer.