It's nearly two years since I last tackled quadratics with a number in front. Recently, though, I stumbled on a slightly different method that's a bit less involved.
I won't say it's easier or better - different methods suit different people, after all - but I like it.
Let's factorise $5x^2 - 2x - 7$, the kind of thing that makes some students turn the page. The trick is to do a sneaky bit of variable-changing under the bonnet - details at the end of the post - which looks very much like moving the 5 to the end. What I mean by that is, you divide the $x^2$ by 5 and multiply the 7 by 5 to get a new expression (changing all the $x$s to $y$s for the sake of good form):
$y^2 -2y - 35$
This looks, and feels, completely non-kosher, but don't worry: it comes out fine. On the plus side, that's easy to factorise: $(y-7)(y+5)$.
The big minus: it doesn't multiply out to what we started with; we need to have a 5 with one of the $x$s. So, what we do is, we divide it out of the right-hand $5$ and multiply it back in the left-hand $x$ - and change the $y$s back into $x$s - making it $(5x -7)(x+1)$.
Now, that does multiply out to the right thing.
Similarly, to sort out $6x^2 + x - 2$, you would convert it into $y^2 + y - 12 = (y-3)(y+4)$, and move a three from left to right and a two from right to left (tricky!) to get $(2x -1)(3x + 2)$ - which multiplies out nicely.
That's a good question. Very well presented. Thank you for asking. I don't have a nice, simple explanation, or else I'd have asked the Mathematical Ninja to show you. I do have an observation, though:
$(ax + b)(cx + d) = ac x^2 + (ad + bc)x + bd$
$(y + bc)(y + ad) = y^2 + (ad + bc)y + abcd$
By shifting the $ac$ - the number in front of the $x^2$ - to the end of the expression, you end up with a factorisation in $y$ that permits switching back to the proper factorisation in $x$ - but I don't have a solid, 'here's why it happens, obviously' explanation. If you know why it works... do let me know!
* Thanks to @S3yM5n for pointing out a typo.