Dear Uncle Colin,

How would you go about factorising $6x^2 - xy - y^2 + 7x - y + 2$?

- Argh! Getting Nowhere. Expression Simplification Impossible.

Hi, AGNESI, and thanks for your message!

I have, historically, not been a fan of these. However, I’ve recently come across a method that makes a bit of sense to me, and I figured it was worth sharing. (I have a second method in case you don’t like the first.)

The plan is:

- Consider what happens if you let each variable in turn be zero
- Combine the resulting factors as best you can.

### A bit at a time

If $x=0$, we have $-y^2 - y + 2$, which I would rewrite as $-(y^2 + y - 2)$ so I don’t lose a minus sign, then as $(y+2)(1-y)$.

If $y=0$, we have $6x^2 + 7x + 2$, which is $(3x + 2)(2x+1)$.

If $x \gg 0$ and $y \gg 0$, we have $6x^2 - xy - y^2$, which is $(3x + y)(2x-y)$.

Now we can combine the brackets: $3x+y$, $3x + 2$ and $y+2$ look like they fit together, so we can make them $(3x+y+2)$.

What’s left over? $2x - y$, $2x+1$ and $-y+1$ also fit together, so the other factor is $(2x - y + 1)$.

We should probably check that, but I’m in a ‘Wolfram|Alpha says it’s right’ mood rather than a ‘multiply it all out’ one.

### Another option

You could also start by factorising *just* the quadratic terms: again, $6x^2 - xy - y^2 = (3x+y)(2x-y)$. To get the linear and constant terms, we will need to add a constant to each of these brackets, making $(3x+y+a)(2x-y+b)$.

Then set up a multiplication grid (other expansion methods are available):

× |
$3x$ |
$y$ |
$a$ |

$2x$ |
$6x^2$ |
$2xy$ |
$2ax$ |

$-y$ |
$-3xy$ |
$-y^2$ |
$-ay$ |

$b$ |
$3bx$ |
$yb$ |
$ab$ |

Then, comparing coefficients, we know that $ab=2$, $2a + 3b = 7$ and $b -a = -1$. These are satisfied by $a=2$ and $b=1$, so the factorisation is $(3x + y + 2)(2x - y + 1)$, as before.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.