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 5x2−2x−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 x2 by 5 and multiply the 7 by 5 to get a new expression (changing all the xs to ys for the sake of good form):

y2−2y−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 xs. 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 ys back into xs - making it (5x−7)(x+1).

Now, that does multiply out to the right thing.

Similarly, to sort out 6x2+x−2, you would convert it into y2+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.

But WHY, Colin? Why does it work?

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)=acx2+(ad+bc)x+bd (y+bc)(y+ad)=y2+(ad+bc)y+abcd

By shifting the ac - the number in front of the x2 - 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 [twit handle = ‘S3yM5n’] for pointing out a typo.