Factoring Quadratics: Asterisk Method I'm teaching my students about factoring quadratics.  We've done GCF, difference of two squares, squared binomials, and grouping.  One of my colleagues then found this asterisk method on line.
It's basically the grouping method but presented in a little different light.  Has anyone ever seen this or used it?  My bosses would prefer learning the guess and check method, or the grouping method, but if you factor out a CGF before, this method works for all of them.  I would like my students to be able to recognize the special cases and not have to resort to any work if they see $a^2x^2-c^2$ or if they see $(ax)^2+2abx+b^2$, but if they are struggling with those and can grasp this method, would anyone here use this method?
 A: It's a pretty slick method, which will solve most of the same problems that factoring by grouping will solve. Using the example in the video, we can instead proceed as follows: $$\begin{align}12x^2-5x-2 &= 12x^2+(3-8)x-2\\ &= 12x^2+3x-8x-2\\ &= 3x(4x+1)-2(4x+1)\\ &= (4x+1)(3x-2).\end{align}$$
I tend to prefer completing the square for its general utility, but for "nicely factorable" trinomials, the asterisk method works just fine.

Added: As you point out, the method will fail for differences of squares with a common factor, unless you pull out the GCF first. That is a drawback to this method, as opposed to factoring by grouping, which does not require us to pull out the GCF first.
A: At How to factor the quadratic polynomial $2x^2-5xy-y^2$? 
I prove that, for $a x^2 + b x + c,$ with discriminant $\Delta = b^2 - 4 a c,$ the given quadratic factors if and only if $\Delta$ is a (non-negative) square. If so, we are guaranteed to be able to factor as
$$  a x^2 + b x + c = (a_1 x + c_1)(a_2 x + c_2)    $$ by beginning with
$$ a_1 = \gcd \left( \; a, \; \frac{b + \sqrt \Delta}{2} \right),  $$ next taking $a_2 = a / a_1.$
Maybe you can do something with that. 
EEDDDDIIIITTTT:
$$  a x^2 + b x  + c  =  \; \left(a_1x+ \left( \frac{b - \sqrt \Delta}{2a_2} \right)  \right) \; \; \left(a_2x+ \left( \frac{b + \sqrt \Delta}{2a_1} \right)  \right)   \;    $$ in integers.
$$ 24929 x^2 + 15966 x - 10403.  $$
$$ \Delta = 15966^2 - 4 \cdot 24929 \cdot(-10403) = 1292258704   $$
$$ \sqrt \Delta = 35948 $$
$$ \frac{b + \sqrt \Delta}{2} = 25957 $$ 
$$ \gcd (24929, 25957) = \gcd ( 24929, 1028) = \gcd (  1028, 257) = 257.  $$
$$ a_1 = 257.$$
$$ a_2 = 24929/257 = 97.$$ 
$$ c_1 = \frac{b - \sqrt \Delta}{2a_2} = \frac{15966 - 35948}{2 \cdot 97} = -103 $$
$$ c_2 = c / c_1 = 101. $$
OR 
$$ c_2 = \frac{b + \sqrt \Delta}{2 a_1} = \frac{25957}{257} = 101, \; \; \; c_1 = c / c_2 = -10403 / 101 = -103. $$ 
$$ 24929 x^2 + 15966 x - 10403 = (257 x - 103)(97 x + 101).  $$
