Why $ax^2+bx+c = a(x-r)(x-s)$, where $r$, $s$ are the roots?

Question

When I was reading about math, I came across the following -

Suppose the roots of the quadratic $ax^2+bx+c$ are $r$ and $s$.

Then $ax^2+bx+c = a(x-r)(x-s)$ for all values of $x$.

Is there any way to prove the above statement? I tried substituting for a few values and it worked but I couldn't prove that it works for all values of $x$.

Answer 1

You need here two lemmas.

1. In the polynomial ring $\mathbb{R}[x]$ there is a division with residues.

2. if $s$ is a root of p(x) then $p(x)=g(x)(x-s)$.

Can you take it from here?

Answer 2

Just prove that $r = \frac{-b+\sqrt{\Delta}}{2a} $ and $s = \frac{-b-\sqrt{\Delta}}{2a}$, where $\Delta = b^2-4ac$. Then your solution comes on its own.
To do this you have to put your expression in the canonic form : $$ ax^2+bx+c = a(x^2+\frac{b}{a}x+\frac{c}{a}) $$ $$ = a[(x+\frac{b}{2a})^2-\frac{b^2-4ac}{a}] $$ and I think you can figure out the end of it, supposing that you effectively have two solutions, which implies that the second member of the substraction is positive...

Well it's a bit long but works