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We know that we can expand the expression \(2 (5q + 4)\) to give \(10q + 8\)

Expanding and factorising are inverse operations.

This means that we can factorise \(10q + 8\) to give \(2 (5q + 4)\)

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

Factorise \(15 - 12m\)

Solution

Look for a common factor and place this outside a bracket.

Both 15 and 12 can be divided by 3.

3 goes outside the bracket.

\(15 - 12m = 3 (?)\)

Now, work out what needs to go inside the bracket.

\(15 ÷ 3 = 5\) and \(12 ÷ 3 = 4\)

Keep the sign between the terms the same.

Answer

\(15 - 12m = 3 (5 - 4m)\)

Check your work out by expanding.

\(3 (5 - 4m) = 15 - 12m\)

Factorise \(20n - 45\)

Solution

Look for a common factor and place this outside a bracket

The highest common factor of 20 and 45 is 5

\(20n - 45 = 5 (?)\)

Work out what needs to go inside the bracket

\(20 ÷ 5 = 4\) and \(45 ÷ 5 = 9\)

\(20n - 45 =5 (4n - 9)\)

Keep the sign between the terms the same

Answer

\(5 (4n - 9)\)

Check by expanding: \(5 (4n - 9) = 20n - 45\)

Example

Factorise this expression fully \(12 - 28p\)

Solution

A common factor of 12 and 28 is 2.

\(12 - 28p = 2 (6 - 12p)\)

While this statement is true, the expression has not be fully factorised.

To do this, the common factor outside the bracket must be the highest common factor and the numbers left in the bracket should not have any common factors.

The highest common factor of 12 and 28 is 4.

Answer

\(12 - 28p = 4 (3 - 7p)\)

Check by expanding: \(4 (3 - 7p) = 12 - 28p\)