TITLE: How can we prove that a quadratic equation has at most 2 roots?
QUESTION [5 upvotes]: A quad equation can be factored into two factors containing $x $, but how can we prove that there no other sets of different factors yielding OTHER VALUES OF $X $?

REPLY [2 votes]: Let $\,f(x)\,$ is a polynomial of degree $2$ with coefficients in a field or domain $\,F$ (e.g. $\Bbb Q,\Bbb R,\Bbb C)$ and suppose that $\,f\,$ has $\,2\,$ distinct roots $\,a,b.\,$ By the Bifactor Theorem below we deduce that $\,f(x) = c(x\!-\!a)(x\!-\!b)\,$ for $\,c\in F.\,$ Thus if $\,d\neq a,b\,$ then $\,f(d) = c(d\!-\!a)(d\!-\!b)\ne 0\,$ since each factor is $\ne 0\,$ (recall $\,x,y\ne 0\,\Rightarrow\,xy\ne 0\,$ in a domain). So $\,f\,$ has at most $2$ distinct roots.
Bifactor Theorem $\ $ Suppose that $\rm\,a,b\,$ are elements of a domian $\rm\,F\,$ and $\rm\:f\in F[x],\,$ i.e. $\rm\,f\,$ is a polynomial with coefficients in $\rm\,F.\,$ If $\rm\ \color{#C00}{a\ne b}\ $ are elements of $\rm\,F\,$ then
$$\rm f(a) = 0 = f(b)\ \iff\ f\, =\, (x\!-\!a)(x\!-\!b)\ h\ \ for\ \ some\ \ h\in F[x]$$
Proof $\,\ (\Leftarrow)\,$ clear. $\ (\Rightarrow)\ $ Applying  Factor Theorem twice, while canceling $\rm\: \color{#C00}{a\!-\!b\ne 0},$
$$\begin{eqnarray}\rm\:f(b)= 0 &\ \Rightarrow\ &\rm f(x)\, =\, (x\!-\!b)\,g(x)\ \ for\ \ some\ \ g\in F[x]\\
\rm f(a) = (\color{#C00}{a\!-\!b})\,g(a) = 0 &\Rightarrow&\rm g(a)\, =\, 0\,\ \Rightarrow\,\ g(x) \,=\, (x\!-\!a)\,h(x)\ \ for\ \ some\ \ h\in F[x]\\
&\Rightarrow&\rm f(x)\, =\, (x\!-\!b)\,g(x) \,=\, (x\!-\!b)(x\!-\!a)\,h(x)\end{eqnarray}$$