Written by Colin+ in quadratics.

A student asks:

I'm currently preparing for my GCSE mocks. I started looking over quadratic expressions and factorisation and it just blew me away - I get stuck trying to work it out with negatives!

First things first: thanks for asking for help. You're not alone: factorising quadratics can be tricky, especially if you take it too quickly. Here's how I'd recommend approaching your revision. (I'm only going to deal with quadratics that don't have a number in front of $x^2$ - although we know a trick about those):

You need to make sure you're hot on your negative numbers, or else you're going to struggle in several bits of the GCSE (not just quadratics). You've probably learned 'helpful' mnemonics like 'a minus and a minus make a plus', which are probably more harmful than anything else.

I'll write a proper guide to negative numbers another time, but you can practise to your heart's content here. If you get one wrong, see if you can figure out why - and work out your own ways to remember the rules.

I don't care whether you use crab claws, a smiley face, a grid, or some other method, as long as it gets you the right result. You need to be able to see something like $(x-8)(x + 6)$ and say: "That's $x^2 - 8x + 6x - 48$" without thinking too hard, and then turn that into $x^2 - 2x - 48$.

Tip: the $x^2$ isn't going to change in the last step, and nor is the $-48$; you just need to say "that's down 8 $x$s and up 6, which leaves me down 2 $x$s- which is $-2x$.

Notice that the final number is $-8 \times 6$ - it's always what you'd get if you ignored the $x$. The middle number, meanwhile is $-8 + 6$ - the sum of the numbers.

Practise expanding quadratics here.

So, factorising is the same thing in the other direction: it's taking something like $x^2 -2x - 48$ and turning it into $(x-8)(x+6)$ - which means you can always check your answer by multiplying out the brackets again.

The trick is to find the pair of numbers that go at the end of the brackets.

Let's say you're factorising something new: $x^2 - 13x + 36$. I would start by listing the factors of $+36$ and see what each pair adds up to:

$a$ | $b$ | $ab$ | $a+b$ |
---|---|---|---|

1 | 36 | 36 | 37 |

2 | 18 | 36 | 20 |

3 | 12 | 36 | 15 |

4 | 9 | 36 | 13 |

6 | 6 | 36 | 12 |

-1 | -36 | 36 | -37 |

-2 | -18 | 36 | -20 |

-3 | -12 | 36 | -15 |

-4 | -9 | 36 | -13 |

-6 | -6 | 36 | -12 |

You want them to add up to $-13$, so the last-but-one line (with $-4$ and $-9$) is the one you want. These numbers just drop into the bracket, so you get:

$(x-4)(x-9)$

If you multiply that out, you get: $x^2 - 4x - 9x + 36 = x^2 - 13x + 36$, which is what you needed!

After a while, you'll start noticing shortcuts which help you get to the answer more quickly. That's great - especially if they're ones you figure out yourself. Those are much easier to remember than whatever the book or the teacher tells you.

Also: keep practicing. Once you've done a hundred or so, spread over a few weeks, you'll be able to do them in your sleep.

Good luck with your studies!

## Paul

Hello Colin. Do you have a similar quadratic method if x squared is greater than one?

## Colin

Hi, Paul, thanks for your question! In fact, I have TWO methods. I suggest having a look at http://www.flyingcoloursmaths.co.uk/category/algebra/quadratics-algebra/ and picking the one you like best 🙂