factoring a quadratic equation I know that if a quadratic equation has roots s and t, you can factor it as (x-s)*(x-t) and I used that property countless times, but I have never quite understood why that should be. Any clarification and/or proof of this would be appreciated.
I suspect that st = c/a. Is there any way how to see this?
 A: Let $t$ be a root of the polynomial $f(x)=0$, so that $f(t)=0$.
Now divide $f(t)$ by $x-t$ obtaining $f(x)=g(x)(x-t)+r$ where $r$ is a constant.
If you don't know the method of division, note that $x^n=x^{n-1}(x-t)+tx^{n-1}$, which we can apply until $n=1$, so that we eventually obtain a constant remainder.
Now we note that, with $x=t$ we have $0=f(t)=g(t)(t-t)+a=a$, so that $a=0$ and $x-t$ is a factor of $f(x)$.
Therefore if $t$ is a root of $f(t)=0$ then $x-t$ is a factor of $f(x)$ - and that applies whatever degree $f(x)$ has. Obviously a quadratic polynomial has only two linear factors.
If $f(x)=0$ has roots $s,t$ we can write $$f(x)=ax^2+bx+c=a(x-s)(x-t)=ax^2-a(s+t)x+ast$$
Equating coefficients we get $s+t=-\frac ba$ and $st=\frac ca$
A: Let $\,ax^2+bx+c=0\;$ be our quadratic equation ($\;a\neq 0\;$) and let $\;r,s\;$ be its roots, then
$$r,s=\frac{-b\pm\sqrt{\Delta}}{2a}\;,\;\;\Delta:=b^2-4ac$$
and thus we get (watch it: it's slightly different of what you wrote!):
$$a(x-r)(x-s)=a\left(x-\frac{-b-\sqrt{\Delta}}{2a}\right)\left(x-\frac{-b+\sqrt{\Delta}}{2a}\right)=$$
$$=a\left(x^2-\frac{\left[-b-\sqrt\Delta-b+\sqrt\Delta\right]}{2a}x+\frac{b^2-\Delta}{4a^2}\right)=a\left(x^2+\frac bax+\frac ca\right)=ax^2+bx+c$$
