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

You need here two lemmas.


*

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

*if $s$ is a root of p(x) then $p(x)=g(x)(x-s)$.
Can you take it from here? 
