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.
That AC-method reduces to factoring a polynomial that is $\,\rm\color{#c00}{monic}\,$ (lead coeff $\color{#c00}{=1})$ as follows
$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\!\!\rm\: \color{#c00}{X^2} + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\overbrace{ac,}^{\rm\qquad\ \ \ \ \ {\bf\large\ \ AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \ X = a\:x \\
\end{eqnarray}$$
In your case
$$ {\begin{eqnarray}
f \, &\,=\,& \ \ \, 6 x^2+\ 11\ x\,\ -\ \ 35\\
\Rightarrow\,\ 6f\, &\,=\,&\!\,\ (6x)^2\! +11(6x)-210\\
&\,=\,& \ \ \ \color{#c00}{X^2}+\, 11\ X\,\ -\ 210,\,\ \ X\, =\, 6x\\
&\,=\,& \ \ (X+21)\ (X-\,10)\\
&\,=\,& \ (6x+21)\,(6x-10)\\
\Rightarrow\ \ f\:=\: \color{#0a0}{6^{-1}}\,(6f)\, &\,=\,& \ (2x+\,\ 7)\ (3x\,-5)\\
\end{eqnarray}}$$
In the final step we cancel $\,\color{#0a0}6\,$ by cancelling $\,3\,$ from the first factor, and $\,2\,$ from the second.
If we denote our factoring algorithm by $\,\cal F,\,$ then the above transformation is simply
$$\cal F f\, = a^{-1}\cal F\, a\,f\quad\,$$
Thus we've transformed by $ $ conjugation $\,\ \cal F = a^{-1} \cal F\, a\ \,$ the problem of factoring non-monic polynomials into the simpler problem of factoring monic polynomials. The same idea also works for higher degree polynomials, see this answer, which also gives links to closely-related ring-theoretic topics.
Best Answer
Short answer: You have to remember, grouping is simply splitting the middle term so that it can be grouped:
Take the expression $ 3x^2+10x+8 $ The terms have no common factors, but if you just split the middle term right: $$3x^2+4x+6x+8$$ You can suddenly "group" the expression like such: $$x(3x+4)+2(3x+4)$$ Woah - Magic! Now we have $(x+2)(3x+4)$!
Long answer: Continuing off Gerry Myerson's answer:
Given the expression: $ ax^2+bx+c $ You could rewrite it as such: $(dx+e)(fx+g)$
Which proves the following is true (multiply it out): $$ ax^2+bx+c=dfx^2+(dg+ef)x+eg $$ What you want to do first is to find $dg$ and $ef$. Since you already know that $df = a$, and $eg =c$, all you have to do is factor $dfeg$, or $ac$, into $dg$ and $ef$ (remember, $dg+ef=b$). Once you have found $dg$ and $ef$, you can rewrite the expression as such: $$dfx^2+dgx+efx+eg $$ Now, you can "group" the equation into two factorable terms: $$dfx^2+dgx = dx(fx+g)$$$$ efx+eg=e(fx+g)$$ Yay! Now your expression is simply: $$dx(fx+eg)+e(fx+eg)$$$$=(dx+e)(fx+eg)$$