Until about three years ago, I had literally no idea how to factorise nasty quadratics. I would turn straight to the quadratic formula, go bing bang boom and say ‘there, job done.’ This was a very effective short-cut – I got a long way with my ignorance – but I’m faintly embarrassed about it now. Especially since I was taught how to do it properly by two of my students (Kat and Cos – to whom I’m still grateful.)
Let’s start with a simple-ish quadratic expression, such as $x^2 -9x + 18$. The way to factorise is to find two numbers that multiply together to make 18 but add to make -9. Eighteen doesn’t have all that many factor pairs – (1, 18), (2, 9), (3,6) and their negative counterparts. The one we’re after is (-3, -6), which just drop into brackets with the $x$s to make $(x-3)(x-6)$. If you multiply that out, you get back to the original equation. So far so simple.
But when you have a number in front of the $x^2$, it’s a bit trickier. How about $2x^2 + 5x – 3$?
The method’s quite similar – except that the number you’re interested in is the first number times the last one ($2 \times -3 = -6$). The possible factors are (1, -6), (2, -3) and the negatives – so to get 5, I want (-1, 6). Here comes the clever bit!
You’re going to change the middle number (with the $x$) into the numbers you just found – so you rewrite the expression as $2x^2 – x + 6x – 3$.
Factorise the two halves of the equation to get $x(2x -1) + 3(2x -1)$, and glomp them together to get $(x + 3)(2x – 1)$. Finished!
One more example: $6x^2 + 5x – 6$.
You’re interested in $6 \times -6 = -36$, which has plenty of factors: (1,-36), (2,-18), (3,-12),(4,-9) and (6,-6) – and their negative friends. So (-4, 9) will give me 5 in the middle.
That leaves me with $6x^2 – 4x + 9x – 6$, which factorises as $2x(3x – 2) + 3(3x – 2)$, or $(2x + 3)(3x – 2)$.
Try it with these!
(If you get stuck, try using the formula and see where it’s gone pear-shaped!)