I remember learning this method of factoring quadratics in middle school or high school, but looking for a name or more information on it leads me to dead ends.
Given:
$ax^2+bx+c=0$
$d*e=a*c$
$d+e=b$
Then the factorization of the quadratic is:
$(x+\frac{e}{a})*(x+\frac{d}{a})$
Proof:
$(x+\frac{e}{a})*(x+\frac{d}{a})=0$
$x^2+\frac{ex+dx}{a}+\frac{ed}{a^2}=0$
$x^2+\frac{x(e+d)}{a}+\frac{ed}{a^2}=0$
Via substitution of the given above:
$x^2+\frac{bx}{a}+\frac{ac}{a^2}=0$
$x^2+\frac{bx}{a}+\frac{c}{a}=0$
$a*(x^2+\frac{bx}{a}+\frac{c}{a})=a*(0)$
$ax^2+bx+c=0$
This is method is known to me as middle term factor.
Lets take an example $f(x)=x^2+6x+8$. We have to find two numbers such that their sum is their product is $8$ and the sum is $6$. So, factors are $(x+4)(x+2)$.
In general, $ax^2+bx-c$ here constant term $ac$ is negative so we have to find two numbers such that their difference is $b$ and the product is $ac$.
Sometimes finding what to add or subtract might be difficult in that case we can use quadratic formula $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$. You will get two solutions from here $x=\alpha,\beta$. Hence your required factors will be $(x-\alpha)(x-\beta)$.