Proof related to factorization of a quadratic equation

Question

Can we prove or disprove this given statement: "If a, b, and c are integers with a≠0, and the roots of the equation $$ax^2 +bx+c=0$$ are rational, then the equation can be factored as $$(ax+m)(x+n)$$ where m is an integer and n is a rational number, and this factorization is unique."

Answer 1

The solutions to your equations are

$$x_1=\frac{-b+ \sqrt{b^2-4ac}}{2a} \qquad \text{and} \qquad x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$.

Hence the solutions are rational if and only if $b^2-4ac$ is a perfect square, which means $-b\pm \sqrt{b^2-4ac}$ is an integer. Then our polynomial factories as

$$ax^2+bx+c=a(x-x_1)(x-x_2)=(ax-ax_1)(x-x_2)=(x-x_1)(ax-ax_2)$$

Since $b^2-4ac=k^2$, if $k$ is odd, then $k^2$ is odd too and this means $b^2$ is odd, so that $b$ is odd. Then $-b\pm k$ is always even. Instead, if $k$ is even, then $k^2$ is divisible by $4$, so that $b^2$ is divisible by $4$ and so $b$ is even. Then $-b\pm k$ is always even. We have obtained

$$ \frac{-b+ \sqrt{b^2-4ac}}{2}$$

is always an integer number. This means that we can choose

$$m:=-ax_1\in \mathbb N \qquad \text{and} \qquad n:=-x_2$$

or

$$m:=-ax_2\in \mathbb N \qquad \text{and} \qquad n:=-x_1$$

Hence that polynomial can be always factorised as you want but the factorisation is not unique. In particular, there are two possibile choices for such factorisation.

Answer 2

The factorization is not unique so the statement can be disproven. For example take $2x^2+6x+4$. We can factor this polynomial as either $(2x+2)(x+2)$ or $(2x+4)(x+1)$, showing that the factorization is not unique.