Proving $ax^2+bx+c=a(x-r_1)(x-r_2)$, where the $r_i$ are the roots

Question

A very simple question. Saw this formula many places earlier, but how do we prove it? $$ax^2+bx+c=a(x-r_1)(x-r_2)$$ Where $r_1$ and $r_2$ are the roots of the quadratic.

P.S.- I have seen a "proof" using Vieta's formulas, but Vieta's formula itself requires this fact in its proof.

Answer 1

For quadratic equation $$0=ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2},$$ its roots are $$r_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ Depending on the sign of $b^2-4ac$, the roots could be distinct reals, a multiple real, or distinct complexes. We actually do not need the fundamental theorem of algebra here, which is usually used for equations of higher orders (e.g., order $5$ or more).

Then we can evaluate it directly \begin{align*} a(x-r_1)(x-r_2)&=a[x^2-(r_1+r_2)x+r_1r_2] \\ &=a\left[x^2-\left(-\frac{b}{a}\right)x+\frac{b^2-(b^2-4ac)}{4a^2}\right] \\ &=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=ax^2+bx+c. \end{align*} I guess people are overthinking about this question.

Answer 2

I don't know if that's what you want, but try factoring $ax^2+bx+c$ and you will have expressions for the roots.

Answer 3

What do you define as "roots" of the quadratic? If they are precisely the two (not necessarily distinct) values that make the quadratic zero, then it's as trivial as setting the factorised expression to zero and deducing that one or the other linear factor has to be zero. The two roots of the LHS correspond in some order to the two of the RHS, giving you the proof.

As to the fact there are two roots (including the case of a repeated root) of a quadratic, you have to use the Fundamental Theorem of Algebra for that, and that is a fairly deep result.