When factorising quadratics of the form x²+bx+c, the usual procedure is to write it like this:

(x+..)(x+..)

and then think of 2 numbers which multiply together to give c and add up to give b. So, for example:

x²+x-12

factorises to

(x+4)(x-3),

because 4 * -3 = -12 and 4-3=1. Possible combinations which multiply together to give 12 are 4 and 3, 6 and 2 and 12 and 1. The only pair with a difference of 1 is 4 and 3 and as we are looking for -12, if we make the 3 negative, then it all works.

Things get harder when the quadratic is of the form ax²+bx+c, where a ≠ 1.

The trick is to add another step. First, find 2 numbers which multiply together to give ac and add up to give b. Then rewrite the equation, splitting the bx term. Once you have done this you will have 4 terms, which can be grouped and factorised in separate pairs. At this stage there should be a common bracket, which allows you to factorise fully. This sounds complicated, and is easier to understand with an example. Lets look at:

6x²+27x+12.

So, ac = 72 and b = 27.

**Step 1.** Find 2 numbers which when multiplied together make 72 and add up to 27. Possible pairs which multiply together to make 72 are 4 and 18, 8 and 9, 6 and 12 and 3 and 24. The only pair which adds up to 27 though is 3 and 24.

**Step2.** Rewrite the equation as 6x²+3x+24x+12.

**Step3.** Factorise the first and second terms as 3x(2x+1) and the third and fourth terms as 12(2x+1), giving 3x(2x+1)+12(2x+1) .

**Step4.** There is a common bracket of (2x+1), so factorise fully to (3x+12)(2x+1).