Factorising

What Does Factorising Mean?

In simple terms, factorising means putting back into brackets. If you think of expanding as spreading out the terms by multiplying, factorising is the opposite: it's about finding the common factors and grouping the terms back together inside a bracket.

How to Factorise

Identify the Highest Common Factor (HCF):

Look for the highest common factors in each term. These could be numbers, letters, or both.

Place the Common Factors Outside the Bracket:

Once identified, place the HCF outside the bracket.

Write What's Left Inside the Bracket:

For each term, divide by the HCF to find what's left inside the bracket.
Ask yourself: What do I need to multiply the term outside the bracket by to get my original term?

Ensure No More Common Factors:

Double-check that there are no further common factors inside the bracket.

Check Your Work:

Quickly expand the bracket to ensure that the factorisation is correct. This is a good habit that takes just a moment and ensures you've done it right.

Understanding Factors

The key to successful factorising is understanding factors. Sometimes it helps to write out each term in full to easily spot the factors.

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Examples:

$12a$ can be broken down into $12×a.$
$6y²$ can be broken down into $6×y×y$ .
$7pq²$ can be broken down into $7×p×q×q.$

Worked Examples

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Example 1: Factorising a Simple Expression

Problem: Factorise $7a+21$ .

Solution:

Step 1: Identify the common factors.
Numbers: The HCF of 7 and 21 is 7.
Letters: There are no letters common in both terms.
Step 2: Place $7$ outside the bracket.

7(?+?)

Step 3: Determine what's left inside the bracket.

7×?=7a→a

7×?=21→3

7×a=7a \quad \text{and} \quad 7×3=21

Final Answer:

7a+21=7(a+3)

Check: Expanding $7(a+3)$ gives $7a+21$ , so the factorisation is correct.

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Example 2: Factorising with Variables

Problem: Factorise $10p+15pq$ .

Solution:

Step 1: Identify the common factors.
Numbers: The HCF of 10 and 15 is 5.
Letters: The letter p is common in both terms.
Step 2: Place $5p$ outside the bracket.

5p(?+?)

Step 3: Determine what's left inside the bracket.

5p×?=10p→2

5p×?=15pq→3q

5p×2=10p \quad \text{and} \quad 5p×3q=15pq

Final Answer:

10p+15pq=5p(2+3q)

Check: Expanding $5p(2+3q)$ gives $10p+15pq$ , so the factorisation is correct.

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Example 3: Factorising with Multiple Variables

Problem: Factorise $24c^2+16c$ .

Solution:

Step 1: Identify the common factors.
Numbers: The HCF of 24 and 16 is 8.
Letters: $c2$ and $c$ share a common factor of c.
Step 2: Place $8c$ outside the bracket.

8c(?+?)

Step 3: Determine what's left inside the bracket.

8c×?=24c^2→3c

8c×?=16c→2

8c×3c=24c² \quad \text{and} \quad 8c×2=16c

Final Answer:

24c²+16c=8c(3c+2)

Check: Expanding $8c(3c+2)$ gives $24c²+16c$ , so the factorisation is correct.

Important Note: Avoiding Common Mistakes

A very common mistake when factorising is not taking out the highest common factor. Let's explore what happens if we miss the highest common factor:

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Example with a Mistake:

Imagine we thought the highest common factor of $24$ and $16$ was $2$ instead of $8$ .
Numbers: $HCF=2$
Letters: $c²$ and $c$ share a common factor of $c$ .
If we factor out $2c$ :

2c(12c+8)

Expanding $2c(12c+8)$ gives $24c²+16c$ , but notice that the terms inside the bracket $12c+8$ still have a common factor of 4.

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Example 4: Factorising with Multiple Variables Problem: Factorise $18bc−45b²$

Solution:

Step 1: Identify the common factors.
Numbers: The HCF of 18 and 45 is 9.
Letters: The terms share a common factor of b.
Remember: $b^2$ is just $b×b$ and $bc$ is just $b×c$
Step 2: Place $9b$ outside the bracket.

9b(?−?)

Step 3: Determine what's left inside the bracket.

9b×?=18bc→2c

9b×?=45b^2→5b

9b×2c=18bc \quad \text{and} \quad9b×5b=45b²

Final Answer:

18bc−45b²=9b(2c−5b)

Check: Expanding $9b(2c−5b)$ gives $18bc−45b²$ , so the factorisation is correct.

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Example 5: Factorising a Complex Expression

Problem: Factorise $18a²b−6ab+30ab²$ .

Solution:

Step 1: Identify the common factors.
Numbers: The HCF of 18, 6, and 30 is 6.
Letters: The common factor is ab across all terms.
Remember: $a^2b$ is just $a×a×b$ and $ab^2$ is just $a×b×b$
Step 2: Place $6ab$ outside the bracket.

6ab(?−?+?)

Step 3: Determine what's left inside the bracket.

6ab×?=18a^2b→3a

6ab×?=6ab→1

6ab×?=30ab^2→5b

6ab \times 3a = 18a^2b \quad \text{and} \quad 6ab \times (-1) = -6ab \quad \text{and} \quad 6ab \times 5b = 30ab^2

Final Answer:

18a²b−6ab+30ab²=6ab(3a−1+5b)

Check: Expanding 6 $ab(3a−1+5b)$ gives $18a²b−6ab+30ab²$ , so the factorisation is correct.

Factorising Simplified Revision Notes for GCSE OCR Maths

Factorising

What Does Factorising Mean?

How to Factorise

Understanding Factors

Example 1: Factorising a Simple Expression

Example 2: Factorising with Variables

Example 3: Factorising with Multiple Variables

Important Note: Avoiding Common Mistakes

Example 5: Factorising a Complex Expression

500K+ Students Use These Powerful Tools to Master Factorising For their GCSE Exams.

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