On page 7 of “Elliptic Tales: curves, counting and number theory” the authors state that:
$(m^2+1)x^2+(2mb)x+(b^2-1)=0$
Has the factors:
$(m^2+1)(x-\alpha)(x-\beta)=0$
Where $\alpha$ and $\beta$ are the roots. Probably this is simple and I’m just rusty but I really don’t see where this comes from, could someone explain? The book states this as though it is a common feature of quadratics but if so it’s unfamiliar to me.
On
Note that $$(m^2+1)x^2+(2mb)x+(b^2-1)= Ax^2 +Bx +C$$
Where $$A= (m^2+1)$$ and $$B= 2mb$$ and $$C= b^2-1$$
Every quadratic equation $$Ax^2 +Bx +C$$ factors as $$A(x-x_1)(x-x_2)$$ where $x_1$ and $x_2$ may be complex, but we can find them by the quadratic formula. $$x=\frac {-B\pm \sqrt {B^2-4AC}}{2A}$$
This is a general property of quadratic functions and has nothing much to do with this particular quadratic. For any quadratic function $f(x)=ax^2+bx+c$, if $\alpha_1$ and $\alpha_2$ are its roots, then $$f(x)=ax^2+bx+c=a(x-\alpha_1)(x-\alpha_2)$$ Similarly, if $P(x)$ is any polynomial of degree $d$ with leading coefficient $c$, it holds that $$P(x)=c\prod_{k=1}^d (x-\alpha_i)$$ where $\alpha_i$ are the roots of $P(x)$.
See this wikipedia page.