if $ax^2+bx+c$ and $2ax+b$ expressions have a common divisor, then $ax^2+bx+c$ is a full square.

Question

I would like to prove this statement.

Thank you for any idea.

Hint: note that the linear polynomial is the derivative of the quadratic. — lulu, Dec 29 '21 at 13:59
Hint: $,x-r,$ is common divisor implies $r$ is a double root by the Double Root Test. This is a dupe so please delete it once it is clear to you. You should specify the coefficient ring. — Bill Dubuque, Dec 29 '21 at 14:09
@Rivers Yes, but it is common abuse of language that "have a common divisor" means the common divisor is nontrivial, i.e. not a unit (invertible). — Bill Dubuque, Dec 29 '21 at 14:20

score 0 · Answer 1 · answered Dec 29 '21 at 14:28

One way to prove this fact is to use the resultant: https://en.wikipedia.org/wiki/Resultant

The following theorem holds ($R(f,g)$ is the resultant of two polynomials):

Theorem: If $A$ is a UFD, $f,g\in A[x]$ have a common irreducible factor if and only if $R(f,g)=0$.

Clearly, if $f,g\in \mathbb{R}[x]$ have a common root then they have a common irreducible factor (the converse is not true in $\mathbb{R}$, it is true in $\mathbb{C}$, since it is an algebraically closed field), and so their resultant must equal zero: $R(f,g)=0$.

So, let's suppose that $f=ax^2+bx+c\in\mathbb{R}[x]$, with $a\neq 0$, and $f'=2ax+b\in\mathbb{R}[x]$ have a common factor. Their resultant must equal zero:

$$R(f,g)=\det\begin{bmatrix} a & b & c\\ 2a & b & 0\\ 0 & 2a & b \end{bmatrix}=-a(b^2-4ac)=0$$

So $\Delta=b^2-4ac=0$, and thus $f$ is a perfect square.

Edit: As said in the comments, the coefficients ring was not declared. I guessed $f,f'\in\mathbb{R}[x]$.

if $ax^2+bx+c$ and $2ax+b$ expressions have a common divisor, then $ax^2+bx+c$ is a full square.

1 Answers1