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.
Best Answer
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 has to be understood 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:
$$z=\lambda z'_1+(1-\lambda) z'_2= \frac13\left(d-a-2 \lambda d \right)\tag{4}$$
where $\lambda \in \mathbb{R}$.
Now plug relationship (4) into (1).
Remark: a supplementary condition is brought by the fact that two of the roots must lie outside open line segment $(z'_1,z'_2)$, implying that the case $0<\lambda<1$ can occur for at most one root.