Higher GCSE

# Factorising Quadratic Expressions for GCSE Maths – Higher Tier

Following on from the last blog on factorising quadratics, here is a guide to factorising quadratics of the form ax2 + bx + c, where a is greater than 1. This will only be tested on the higher tier GCSE maths exam, so if you’re studying foundation, this blog is of no use to you!

Let’s work through the method with some examples:

Example 1:
Factorise 2x2 + 7x + 6

Firstly, multiply a by c.

a, in this case, is 2, c is 6.

2 × 6 = 12

Now, in a way very similar to the method for normal quadratics, we are looking for two numbers that multiply to get this number (12) and add to get the middle number (7).

These pairs of numbers multiply to get 12.

1 & 12         1 + 12 = 13
2 & 6           2 + 6 = 8
3 & 4           3 + 4 = 7

(Again, you will realise that there are also three other options: –1 & –12, –2 & –6 and –3 & –4. But because our quadratic contains no negative numbers, we can ignore these this time.)

So it’s going to be 3 & 4.

Now, this is where it gets a little bit different.

* Rewrite the quadratic splitting the 7x into the two numbers (3 & 4), like so:

2x2 + 3x + 4x + 6

Draw a line through the middle to split the quadratic in half like so, and factorise each side into one bracket.

If you’re unsure how we got that, we split the quadratic in two, leaving a red part and a blue part.

We factorise the red part and blue part separately:

2x2 + 3x factorises to x(2x + 3)

4x + 6 factorises to 2(2x + 3)

NOTE: the brackets here should always match – it’s (2x + 3) in this case. If they don’t, you have done something wrong.

So our quadratic can be written as…

x(2x + 3) + 2(2x + 3)

which can then be written as…

(x + 2)(2x + 3)

That last step is often the part that confuses people. This usually helps to explain it…

x(2x + 3) + 2(2x + 3)

Let’s imagine (2x + 3) is A. We would have Ax + 2A. This would factorise to A(x + 2).

Then substitute A = (2x + 3) back in, and you get (2x + 3)(x + 2).

And that is the solution. Just as we did with the foundation quadratics, let’s expand it out to double–check…

(x + 2)(2x + 3)       = 2x2 + 4x + 3x + 6
= 2x2 + 7x + 6

It matches, so the answer is correct.

Now, one question you may ask is – what would happen if, at the *, you had split the 7x into 4x + 3x instead of 3x + 4x?

Well, the good news is – it doesn’t matter, you’ll get the same answer. Look:

We get (2x + 3)(x + 2). It’s exactly the same, you just get (x + 2) as the common bracket instead of (2x + 3).

Let’s do another…

Example 2:
Factorise 2x2 + x – 10

Firstly, multiply a by c.

a, in this case, is 2, c is –10.

2 × –10 = –20.

We are looking for two numbers that multiply to get –20 and add to get 1 (remember +x means +1x).

These pairs of numbers multiply to get –20 (there will be more pairs this time because of the minus sign):

–1 & 20                 –1 + 20 = 19
–20 & 1                 –20 + 1 = –19
–2 & 10                 –2 + 10 = 8
–10 & 2                 –10 + 2 = –8
–4 & 5                   –4 + 5 = 1
–5 & 4                   –5 + 4 = –1

So it’s going to be –4 & 5.

Now, rewrite the quadratic splitting the x into the two numbers (–4 & 5)…

Draw a line through the middle to split the quadratic in half, and factorise each side into one bracket.

So the answer is (2x + 5)(x – 2). Let’s expand it out to double–check…

(2x + 5)(x – 2)        = 2x2 + 5x – 4x – 10
= 2x2 + x – 10

This matches the question, so we know the answer is correct.

Here are two further examples.

Example 3:
Factorise 4x2 + 4x – 15

4 × –15 = –60

So we are looking for two numbers that multiply to get –60 and add to get 4.

These pairs multiply to get –60 (this time there are quite a few!):
–1 & 60                 –1 + 60 = 59
–60 & 1                 –60 + 1 = –59
–2 & 30                 –2 + 30 = 28
–30 & 2                 –30 + 2 = –28
–3 & 20                 –3 + 20 = 17
–20 & 3                 –20 + 3 = –17
–4 & 15                 –4 + 15 = 11
–15 & 4                 –15 + 4 = –11
–5 & 12                 –5 + 12 = 7
–12 & 5                 –12 + 5 = –7
–6 & 10                 –6 + 10 = 4
–10 & 6                 –10 + 6 = –4

So it’s going to be –6 & 10. Split the 4x into –6x and 10x…

So the answer is (2x + 5)(2x – 3)

Expand to check…

(2x + 5)(2x – 3)      = 4x2 + 10x – 6x – 15
= 4x2 + 4x – 15

This matches the question, so we know the answer is correct.

Example 4:
Factorise 6x2 – 5x – 4

6 × –4 = –24

So we are looking for two numbers that multiply to get –24 and add to get –5.

These pairs multiply to get –24 (there are quite a few! If you are struggling with all this mental maths, we have written a post with some games you can play to practice your multiplication tables):

–1 & 24                 –1 + 24 = 23
–24 & 1                 –24 + 1 = –23
–2 & 12                 –2 + 12 = 10
–12 & 2                 –12 + 2 = –10
–3 & 8                   –3 + 8 = 5
–8 & 3                   –8 + 3 = –5
–4 & 6                   –4 + 6 = 2
–6 & 4                   –6 + 4 = –2

So it’s going to be –8 & 3. Split it out…

Notice here that (3x – 4) cannot factorise, so we have to put it as 1(3x – 4).

So the answer is (2x + 1)(3x – 4)

Expand to check…

(3x – 4)(2x + 1)      = 6x2 – 8x + 3x – 4
= 6x2 – 5x – 4

This matches the question, so we know the answer is correct.

So there’s your guide to factorising quadratics on the higher tier. You will also of course still need to be able to factorise the quadratics from the foundation tier – if you need a recap on that method, take a look at the previous blog.

If you want someone to work through this with you in person, book a free taster session here.

Have a go at these on your own. Email your answers to sam@metatutor.co.uk or put them in the comments and I will let you know if you are right!

Factorise:
1. 2x2 + 7x + 5
2. 3x2 + 10x + 8
3. 2x2 + 5x – 12
4. 5x2 + 12x + 4
5. 4x2 + 15x – 4
6. 5x2 – 6x – 8
7. 6x2 – 29x – 5
8. 7x2 + 23x + 6
9. 8x2 – 2x – 3
10. 4x2 – 11x + 6

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