Towards the end of a GCSE paper, you're quite frequently asked to simplify an algebraic fraction like:

$\frac{4x^2 + 12x - 7}{2x^2 + 5x - 3}$

Hold back the tears, dear students, hold back the tears. These are easier than they look. There's one thing you need to know: algebraic fractions are happiest when they're in brackets.

If you're a regular reader, you'll know how to put quadratic expressions in brackets - check out this article, or this one if you prefer.

To factorise the top, you convert it to $X^2 + 12X - 28$ (shuttling the 4) and factorise: $(X+14)(X-2)$. You then move a two from the 14 to the opposite $X$, and a two from the -2 to to opposite X to get $(2x - 7)(2x - 1)$. Lovely.

The bottom works much the same way: it becomes $X^2 + 5X - 6$, or $(X+6)(X-1)$. Shuttle a 2 from the -6 to the other X to get $(x-3)(2 x -1)$.

The fraction is now $\frac{(2 x -7)(2 x -1)}{(x-3)(2 x -1)}$. Aha! There's a common factor of $(2 x -1)$ on the top and the bottom, so we can cancel that to get: $frac{2x -7}{x-3}$ - which is our fully-simplified answer.

Once you've done a handful of these, you'll start to get a Pavlovian response to this kind of algebraic fraction, and dive straight in!

One thing to look out for is difference of two squares, which comes up once in a while and catches some students out. But you're smarter than that, right? Right.

* Edited 2014-09-11 for clarity and to fix LaTeX errors.

## 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.

## Shaun

hi how to factorise simplify further

2 2

23 – 177

## Colin

Hi, Shaun,

I’m not sure from the formatting whether you mean $23^2 – 177^2$ or $\frac{2}{23} – \frac{2}{177}$.

If it’s the first, it’s the difference of two squares: $(23 + 177)(23-177) = (200)(-154) = -30,800.$

If it’s the second, it’s $\frac{2 \times 177}{23 \times 177} – \frac{2\times 23}{23 \times 177} = \frac{354}{4071} – \frac{46}{4071} = \frac{308}{4071}$

## Jo

Hi, how do you factorise $\frac{2}{y-1} – \frac{1}{2-y} = \frac{6}{y}$?

## Colin

(I’ve edited your comment to what I think you meant – please let me know if I did it wrong!)

I would multiply everything through by denominators until I had no fractions left.

Stage 1, multiply by $y-1$: $2 – \frac{y-1}{2-y} = \frac{6(y-1)}{y}$

Stage 2, multiply by $2-y$: $2(2-y) – (y-1) = \frac{6(y-1)(2-y)}{y}$

Stage 3, multiply by $y$: $2y(2-y) – y(y-1) = 6(y-1)(2-y)$

Now expand: $(4y – 2y^2) – (y^2 – y) = 6(-y^2 + 3y -2)$

Expand and simplify: $-3y^2 + 5y = -6y^2 + 18y – 12$

Regroup on the left: $3y^2 – 13y + 12 = 0$

And finally factorise: $(3y-4)(y-3) = 0$.