Factorising
In Algebra, factorising is the opposite of expanding.
Prior Knowledge 鈥 M1 Factorising
Factorising expressions
We know that we can expand the expression \(2p (q + 4)\) to give \(2pq + 8p\)
Expanding and Factorising are inverse operations.
This means that we can factorise \(2pq + 8p\) to give \(2p (q + 4)\)
To factorise an expression
- Look for a common factor and place this outside a bracket
- Work out what needs to go inside the bracket to keep the expression correct when multiplied out
- Keep the sign between the terms the same
Example
Factorise \(9m 鈥 12m虏\)
Solution
Look for a common factor and place this outside a bracket
The highest common factor of 9m and 12m虏 is 鈥3m鈥
Think about the numbers and the 鈥榤鈥 terms separately.
'3' is the HCF of 9 and 12
'm' is the HCF of m and m虏
3m goes outside the bracket.
\(15m 鈥 12m虏 = 3m ( ? )\)
Work out what needs to go inside the bracket
Keep the sign between the terms the same
\(15m 鈥 12m虏 = 3m (5 鈥 4m)\)
Answer
\(15m 鈥 12m虏 = 3m (5 鈥 4m)\)
Check by expanding
\(3(5m 鈥 4m) = 15m 鈥 12m虏\)
Question
Factorise \(5tm 鈥 35m\)
Solution:
Look for a common factor and place this outside a bracket
The highest common factor of 5tm and 35m is 5m
\(5tm 鈥 35mn = 5m ( ? )\)
Work out what needs to go inside the bracket
\(5tm 鈥 35m = 5m (t 鈥 7)\)
Answer:
\(5m (t 鈥 7)\)
Check
\(5(t 鈥 7m) = 5t 鈥 35m\)
Example
Factorise this expression fully \(4q 鈥 12q^2\)
Solution
One common factor of 4 and 12 is 2
\(4q 鈥 12q虏 = 2(2q 鈥 6q虏)\)
While this statement is true, the expression has not been fully factorised.
To do this, the common factor outside the bracket must be the highest common factor.
The HCF of \(4\) and \(12\) is \(4\) and the HCF of \(q\) and \(q^2\) is \(q\)
Answer
\(4q 鈥 12q^2 = 4q (1 鈥 3q)\)
Check
\(4q (1 鈥 3q) = 4q 鈥 12q^2\)
Question
Factorise fully \(15x - 4xy - 7x^2\)
Solution:
\(x\) is the only factor common to all three terms
\(15x - 4xy 鈥 7x^2 = x (15 - 4y - 7x)\)
Answer:
\(15x - 4xy 鈥 7x^2 = x (15 - 4y - 7x)\)
Test yourself
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