Following on from the two previous blogs on factorising quadratic for foundation tier maths and higher tier maths factorising, in this blog I will use an example to explain why factorising is useful and how it will help you pass your maths GCSE.

When you were first taught about putting things into brackets, you probably thought it was just something mean mathematicians made up just to put in your maths exams. You probably wondered why algebra would ever be useful. You probably wondered why we bothered doing it. In fact, factorising is an incredibly powerful tool in mathematics. For example, the question below…

On the face of it, this problem looks very difficult. Most GCSE maths students would take one look at it and then turn over the page. But actually, by factorising, we can break this down to something tiny. Like turning a dragon into a chihuahua…

Let’s deal with the first fraction.

Let’s factorise the numerator (x^{2} + 2x – 8).

Going back to the first quadratics blog, we know it is going to look like (x )(x ), we just have to figure out what the two numbers inside the brackets are.

We are looking for two numbers that multiply to get -8 and add to get 2.

These pairs of numbers multiply to get -8:

-1 & 8 -1 + 8 = 7

-8 & 1 -8 + 1 = -7

-4 & 2 -4 + 2 = -2

-2 & 4 -2 + 4 = 2

So it’s going to be -2 & 4.

So it is (x – 2)(x + 4).

Now let’s factorise the denominator (x^{2} + 7x + 12).

We are looking for two numbers that multiply to get 12 and add to get 7.

These pairs of numbers multiply to get 12:

1 & 12 1 + 12 = 13

2 & 6 2 + 6 = 8

3 & 4 3 + 4 = 7

So it is (x + 3)(x + 4).

Our first fraction can now be written as:

You will notice that there is a common factor of (x + 4) on the top and bottom of the fraction. This means we can cancel these terms off. Leaving us with this:

Now, let’s do the same to the second fraction.

Let’s factorise the numerator (x^{2} – 7x + 10).

We are looking for two numbers that multiply to get 10 and add to get -7.

These numbers multiply to get 10:

1 & 10 1 + 10 = 11

2 & 5 2 + 5 = 7

Neither of these work, so we need to look at the negative numbers:

-1 & -10 -1 + -10 = -11

-2 & -5 -2 + -5 = -7

So it is (x – 2)(x – 5).

Now, the denominator (x^{2} – 2x – 15)…

We are looking for two numbers that multiply to get -15 and add to get -2.

These numbers multiply to get -15:

-1 & 15 -1 + 15 = 14

-15 & 1 -15 + 1 = -14

-3 & 5 -3 + 5 = 2

-5 & 3 -5 + 3 = -2

So it is (x – 5)(x + 3).

The second fraction can now be written as:

Again, as before, we have a common factor that we can cancel off:

So, putting these two together leaves us with:

You might already be able to spot what the answer is, but let’s do it the long way just to illustrate the method.

Remember when dividing fractions, we can change the divide sign to a multiply sign, and flip the second fraction (you may remember it as Keep Change Flip)…

Merging these together we get…

Now, you might notice we can cancel off the (x – 2)’s…

Oh, and look, we can also cancel off the (x + 3)’s…

Which, believe it or not, leaves us with an answer of 1.

So we have just found out that…

So that big nasty dragon we had at the start has turned into a cute little chihuahua! All thanks to factorising.

And the really cool (or at least I think it is) thing about this, is that this is true for all values of x. It doesn’t matter what x is – it could be 2, it could be 0.4, it could be 12,345,865. Try any number you can think of – you will always get 1.

If you find this as exciting as I do, then why not give these a try! Put your answers in the comments below or email them to me at sam@metatutor.co.uk and I’ll let you know if you got them right.

If factorising a quadratic expression still doesn’t quite make sense and you need someone to explain this to you in person, book in a free taster session with one of our maths tutors here.

For even more questions like this, try our algebraic fractions worksheet. And if you want to brush up on other aspects of algebra, you can try our algebra worksheet.