In this article, we cover the key factorisation techniques you need to know for Year 9. Test your algebra skills with our questions at the bottom of the page!

Factorisation is the reverse process of expanding and is a powerful tool in algebra at every level of mathematics. It provides us with a way to solve quadratic equations, simplify complicated expressions, and sketch non-linear relationships in Year 10 and beyond.

In factorisation, we want to insert brackets. What makes factorising tricky is that there are many different types. You will need lots of practice to be able to quickly recognise the different types and master the different methods to apply each.

- Stage 5.2: Factorise algebraic expressions by taking out a common algebraic factor (ACMNA230)
- Stage 5.3: Factorise monic and non-monic quadratic expressions (ACMNA269)
- common factors
- grouping in pairs for four-term expressions
- a difference of two squares
- perfect squares
- quadratic trinomials (monic and non-monic)

Students should be familiar with basic algebraic techniques including expanding special binomial products and simple arithmetic. Knowledge of lowest common multiples (LCM) and highest common factors (HCF) will also be required.

This is the simplest form of factorising and involves taking out the highest common factor (HCF) from two or more terms. Note that the HCF may be a term in brackets as well.

Step 1: Find the HCF of all the terms in the expression.

Step 2: Extract the HCF and introduce brackets to form a product.

After the common factor has been taken out, the terms remaining in the brackets should have no other factors in common.

Since factorising is the opposite of expanding, you can always check whether you have factorised correctly by expanding your result and seeing if it matches up with what you started with.

Sometimes, there may not be a HCF for every single term in the algebraic expression. In these instances, we group the terms in pairs so that the first pair of terms have a HCF and the remaining pair of terms have a different HCF. It is important that you group the terms correctly to lead to a successful factorisation.

After extracting the respective HCF from each pair, you will find another common factor. Extract this to produce your final factorised answer.

Factorise the algebraic expression by grouping in pairs

\(2xy+3yz-4x-6z \)Step 1: Regroup the terms such that each pair has a HCF.

\( =(2xy+3yz)-(4x+6z)\)Step 2: Extract the HCF from each pair.

\(=y(2x+3z)-2(2x+3z) \)Step 3: Extract the resulting common factor.

\((2x+3z)(y-2) \)The order that the terms in the brackets are written do not matter. \((a+b)(c+d)=(c+d)(a+b)\). This is an example of the commutative law of multiplication.

There are three special factorising identities that will help you to factorise different types of algebraic expressions. The first is known as the difference of two squares. By expanding, we can show that \((x-y)(x+y)=x{^2}-y{^2} \).

Hence, to factorise the difference of two squares:

\( x{^2}-y{^2}=(x-y)(x+y)\)

Factorise the following algebraic expressions:

(i)\( x{^2}-9y{^2}\)

(ii)\(4x{^2}-81y{^2} \)

Step 1: Rewrite the expression as a difference of two squares.

\( x{^2}-9y{^2}=(x){^2}-(3y){^2}\)

Step 2: Factorise using the rule.

\( (x){^2}-(3y){^2}=(x-3y)(x+3y)\)

Step 1: Rewrite the expression as a difference of two squares.

\(4x{^2}-81y{^2}=(2x){^2}-(9y){^2} \)

Step 2: Factorise using the rule.

\((2x){^2}-(9y){^2}=(2x-9y)(2x+9y) \)

A perfect square is an algebraic product that can be written in the form \( (x+y){^2}\) or \((x-y){^2} \).

When we expand a perfect square, we get the following result:

\((x \pm y){^2}=x{^2} \pm 2xy+y{^2} \)

From this we can see that the middle term \( 2xy\) is twice the product of the numbers \(x \)and \(y \)in the bracket and the first and third terms are perfect squares of them. This identity is what we will be using to factorise perfect squares.

Factorise the following algebraic expressions:

(i) \(a{^2}+6a+9 \)

(ii) \( 16a{^2}+40a+25\)

Step 1: Check whether it is a perfect square.

From the first and third terms we know that \( x=a\) and \( y=3\).

Therefore, \(2xy=2(a)(3)=6a \), which is the middle term in the expression.

Step 2: Factorise:

\( a{^2}+6a+9=(a+3){^2}\)

**Step 1**: Check whether it is a perfect square.

From the first and third terms we know that \( x=4a \) and \(y=5 \).

Therefore, \(2xy=2(4a)(5)=40a \), which is the middle term in the expression.

Hence this expression is a perfect square.

**Step 2**: Factorise:

A quadratic trinomial is an expression of the form \( ax{^2}+bx+c\) where \(a \), and \(b \) are given numbers.

A monic trinomial is when the leading co-efficient, \(a=1 \).

When we expand \((x+ \alpha)(x+ \beta) \), we get \( x{^2}+(\alpha + \beta)x+\alpha \beta\) .

The coefficient of \((\alpha + \beta) \) and the coefficient of the constant is \( \alpha \beta\) .

Hence, to factorise a monic quadratic trinomial, we must reverse the process by finding two numbers whose sum is the coefficient of x and product is the constant term

Factorise the algebraic expression \( x{^2}-5x+6\).

Find two numbers whose sum is -5 and whose product is +6.

The only possible numbers are -3 and -2.

Therefore \(x{^2}-5x+6=(x-3)(x-2) \)

Sometimes it may be necessary to extract a HCF from the expression before factorising the quadratic trinomial using these strategies. For example, the expression \(3x{^2}-15x+18 \) can be factorised by first removing the HCF of 3.

\(3(x{^2}-5x+6)=3(x-3)(x-2) \).

A non-monic quadratic trinomial is an expression of the form \(ax{^2}+bx+c \) where \(a \ne 0 \).

There are three main strategies for factorising these types of expressions:

- The pairing method,
- The fraction method and
- Cross method.

The example here will use the pairing method. To factorise a non-monic quadratic trinomial, find two numbers whose:

- Sum is the coefficient of \(x \)
- Product is the product of the coefficient of \(x{^2} \) and the constant

Factorise the algebraic expression \(3x{^2}+5x+2 \).

Step 1: Find the product of the coefficient of \(x{^2} \) and the constant.

It is 6.

Step 2: Find two numbers whose sum is 5 and whose product is 6.

The only possible numbers are 2 and 3.

Step 3: Use these two numbers to split the middle term and then factorise by grouping in pairs.

\( 3x{^2}+5x+2 = (3x+2)(x+1)\)

All three methods are taught in Matrix theory lessons to expose students to a variety of strategies for factorising non-monic quadratic trinomials. Different schools will teach different methods, but students should select the method that best suits their learning style and practice the strategy until they have mastered it.

Check your factorisation skills with the following 10 exercises!

1. \(4ab{^2}c-6abc{^2}+12a{^2}bc{^2} \)

2. \((a+3b){^2}-9(a+3b)(a-3b) \)

3. \( x{^3}-3x{^2}+2x-6\)

4. \( (3x+4y){^2}-(2x+y){^2}\)

5. \(16x{^4}-81y{^4} \)

6. \(k{^2}-18k+81 \)

7. \(t{^2}-17t-60 \)

8. \( 5m{^2}n-20mn-105n\)

9. \(3x{^2}-11x-20 \)

10. \( 3-10x-8x{^2}\)

1. \(2abc(2b-3c+6ac) \)

2. \(2(a+3b)(15b-4a) \)

3. \((x{^2}+2)(x-3) \)

4. \(5(x+y)(x+3y) \)

5. \((4x{^2}+9y{^2})(2x+3y)(2x-3y) \)

6. \((k-9){^2} \)

7. \((t-20)(t+3) \)

8. \( 5n(m-7)(m+3)\)

9. \((3x+4)(x-5) \)

10. \(-(4x-1)(2x+3) \)

Looking to prepare for your Maths exam? Try our Year 9 Maths Max Series Volume 1: An Exam Preparation Workbook that contains examples and questions on the topics ‘Algebraic Techniques and Surds & Indices’.