I've been reading two algebra textbooks from the early 1900's. Each presents an algorithm for factoring quadratic trinomials when the coefficient of the $x^2$ term does not equal one. The technique converts the problem into one that can be solved in the same manner as solving problems where the coefficient of the $x^2$ term is equal to one (i.e., find factors of $c$ that sum to $b$).
In the first example, the coefficient of the $x^2$ term is a square and its square root is exactly contained in the coefficient of the $x$ term. Here is an example:
I cannot figure out why this method works and there is no additional detail in the book or anywhere else that I've looked that describes this method.
The second method applies when the $x^2$ coefficient is not a square, or is a square but that is not contained exactly in the coefficient of the $x$ term. Here is an example:
Conceptually, I understand how this works because the factors of $b$ should contain the same factors in $a$ and $c$, so finding factors based on a $c$ term that is the product of $a$ and $c$ makes sense. However, I don't see this connection in the first example even though in any quadratic trinomial, the following should be true: $$ (Px + m)(Qx + n) = PQx^2 + (Pn + mQ)x + mn $$
As you can see, the $b$ term contains the factors $P,n,m,Q$ which are the same factors in the $a$ term and the $c$ term.
Can anyone provide a hint as to why the first example works?
In the first case you take an equation of the form $a^2 x^2+a b x+c=0$ and rewrite it as $(ax)^2+b(ax)+c=0$; the substitution $y=ax$ then yields $y^2+by+c=0$. If we instead assume the form $a^2 x^2+b x+c=0$, we first multiply by $a^2$ to get
$$a^4 x^2+a^2 b x+a^2 c=0=(a^2 x)^2+b(a^2 x)+c=0$$
So the substitution $y=a^2 x$ simplifies this to $y^2+by+c=0$. In both cases we've substituted $x=y/A$ for some $A$ to get a monic polynomial. This is typically easier to solve by inspection, since the roots $r,s$ of $$y^2+by+c=(y-r)(r-s)=0$$ will add to $b$ and multiply to $rs$. Once this is done, we back-substitute $y=Ax$ to get $$(Ax-r)(Ax-s)=0$$ which is factorized.