Factorising (OCR GCSE Maths): Revision Notes

Examples:

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

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.

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

• Final Answer:

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

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.

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

• Final Answer:

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

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.

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

• Final Answer:

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

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 $2c$:

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

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.

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

• Final Answer:

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

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.

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

• Final Answer:

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