## What you need to know

### Factorising Quadratics

Quadratics are algebraic expressions of the form

ax^2 + bx + c

Where a, b, and c are all numbers. We’ve seen factorising a bunch of terms into one bracket by taking all the common factors outside the bracket – if you haven’t seen this, you should revise factorising. The following is an example of a quadratic with numbers.

x^2 - x + 12

Where there isn’t a number in front of the x^2 or x like this example, then you know the number is a 1.

To factorise quadratics you need to find:

- two numbers which
__times together to make the constant term__ - and
of x.__add together to make the coefficient__

### Take Note:

Factorising quadratics is a useful technique to learn. Having a good understanding of quadratic equations is key and this topic is the first time you are likely to have come across a quadratic. The following topics which require this key knowledge include:

- Solving quadratics through factorisation
- Solving quadratics using the quadratic formula
- Solving quadratics by completing the square

### Example 1: Factorising Quadratics

Factorise x^2 - x - 12.

We are looking for two numbers which multiply to make -12 and add to make -1. Let’s consider some factor pairs of -12.

-12=(-1)\times 12,\text{ or }(-2)\times 6,\text{ or }(-3)\times 4,\text{ or }(-4)\times 3

We could keep going, but there’s no need because the last pair, -4 and 3, add to make -1. This pair fills both criteria, (as highlighted above) so the factorisation of x^2 - x - 12 is

(x - 4)(x + 3)

Note: You can try expanding the double brackets to check your answer is correct. You should always get your original quadratic equation if you do this correctly.

### Example 2: Factorising Quadratics – The Difference of Two Squares

Factorise x^2 - 9.

There’s no x term in this quadratic. Therefore we’re looking for two numbers which multiply to make -9 and add to make 0.

For 2 numbers to add to make zero, they must be the same number but with different signs. Furthermore, two of the same numbers being multiplied to make -9 must correspond to a square root. The square root of 9 is 3, and indeed, the pair: 3 and -3 satisfy both criteria of adding to zero and multiplying to make -9. Therefore, the factorisation is

(x + 3)(x - 3)

This special case is called the difference of two squares, because it always takes the form of x^2 takeaway some other square number.

Note: So, if you see a quadratic with no x term, then you know that it can be factorised quickly – with the square root of the constant term in both brackets – one negative and one positive.

### Example 3: Factorising Harder Quadratics

Factorise 2x^2 + 7x + 3.

To do this, we must firstly multiply the number in front of x^2 by the constant term, so in this case that is: 2 \times 3 = 6. Then, we should look for two numbers which multiply to make 6, and add to make 7.

6 = 2 \times 3, \text{ or } 1 \times 6

1 and 6 satisfy both criteria. Now we must rewrite the quadratic with the x split into two parts. In other words, because our pair is 1 and 6, we write the 7x term as x + 6x:

2x^2 + x + 6x + 3

Next, we have to treat the quadratic like two separate expressions and factorise the first two terms into their own single bracket, then factorise the last two terms into their own single bracket:

x(2x + 1) + 3(2x + 1)

Note: If you’ve done this correctly, both brackets end up the same as shown above.

This means that we can now take the bracket itself, (2x + 1), out as a factor to get

(2x + 1)(x + 3)

which is the full factorisation of the quadratic.

### Example Questions

1) Factorise a^2 + a - 30

We are looking for two numbers which add to make 1 and multiply to make -30.

The factors of 30 that satisfy theses two requirements are 5 and 6.

Therefore, the full factorisation of a^2 + a - 30 is

(a - 5)(a + 6)

2) Factorise k^2 - 5k + 6

We are looking for two numbers which add to make -5 and multiply to make 6.

The factors of 6 that satisfy theses two requirements are -2 and -3.

Therefore, the full factorisation of k^2 - 5k + 6 is

(k - 2)(k - 3)

3) Factorise x^2 + 7x + 12

We are looking for two numbers which add to make 7 and multiply to make 12.

The factors of 12 that satisfy theses two requirements are 3 and 4.

Therefore, the full factorisation of x^2 + 7x + 12 is,

(x + 3)(x + 4)

4) Factorise 2x^2 - 12x + 10

(HIGHER ONLY)

Here we can take a factor of 2 out so that we have,

2(x^2 - 6x +5)

We are looking for two numbers which add to make -6 and multiply to make 5.

The factors of 5 that satisfy theses two requirements are -1 and -5.

Therefore, the full factorisation of 2x^2 -12x + 10 is,

2(x -1)(x -5)

5) Factorise 4m^2 - 5m - 6

(HIGHER ONLY)

Firstly, 4 \times (-6) = -24. So, we are looking for two numbers which multiply to make -24 and add to make -5. Let’s look at some factor pairs for -24.

-24 = 1 \times (-24) \text{ or } 2 \times (-12)\text{ or } 3 \times (-8),

We can stop here because -8 + 3 = -5 as desired, so -8 and 3 is our pair. So, now we can rewrite the quadratic, splitting up -5m into -8m + 3m:

4m^2 - 8m + 3m - 6

Now, we factorise the first two terms into a single bracket and the last two terms into another single bracket to get

4m(m - 2) + 3(m - 2)

As per the plan, both of the brackets are the same. This means we can take out (m - 2) as a factor of both terms in the expression. This leaves us with

(m - 2)(4m + 3)

Which is the full factorisation of the quadratic.

### Worksheets and Exam Questions

#### (NEW) Factorising Quadratics Exam Style Solutions - MME

Level 4-5#### (NEW) Factorising Harder Quadratics Exam Style Questions - MME

Level 6-7#### Factorising Quadratics - Drill Questions

Level 4-5#### Algebra Quadratics Expand and Factorise - Drill Questions

Level 4-5### Videos

#### Factorising Harder Quadratics Q1

GCSE MATHS#### Factorising Harder Quadratics Q2

GCSE MATHS#### Factorising Harder Quadratics Q3

GCSE MATHS### Learning resources you may be interested in

We have a range of learning resources to compliment our website content perfectly. Check them out below.