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Factorising Linear Expressions

Factorising is expanding in reverse. Instead of multiplying a bracket out, you spot what every term shares and pull it back outside, writing the expression as a product again.

MYP 4Standard MathsAlgebraCriteria A · C · D~10 min read

Written by Fayek Jawad Haider · MYP student

The idea

When you expand \(3(2x + 3)\) you get \(6x + 9\). Factorising runs that backwards: given \(6x + 9\), you recover \(3(2x + 3)\). The two forms are equal; one is a sum, the other a product.

Factorise
To write an expression as a product, by taking a common factor out of every term and placing it in front of a bracket.

A fully factorised answer takes out the highest common factor, the largest thing every term shares. Taking out too little leaves the bracket still factorisable.

Finding the highest common factor

Look at the numbers first. The highest common factor (HCF) of the coefficients is the biggest number that divides all of them exactly.

Worked example

Factorise \(6x + 9\).

1
The coefficients are \(6\) and \(9\). The largest number dividing both is \(3\).
2
Divide each term by \(3\): \(6x \div 3 = 2x\) and \(9 \div 3 = 3\).
3
Write the \(3\) outside, the results inside a bracket: \(3(2x + 3)\).
\(3(2x + 3)\)

Take the highest factor, not just any factor

Writing \(6x + 9\) as \(3(2x + 3)\) is complete. Writing it as \(1(6x + 9)\) is not factorising at all, and the bracket in \(3(2x+3)\) shares no further factor, which is how you know you are finished.

Factors that include letters

If every term also contains the same letter, that letter is part of the common factor too. Take out the number and the shared letters together.

Worked example

Factorise \(6a + 9ab\).

1
The coefficients \(6\) and \(9\) share the factor \(3\). Both terms contain \(a\), so \(a\) comes out too.
2
The common factor is \(3a\). Divide each term by it: \(6a \div 3a = 2\) and \(9ab \div 3a = 3b\).
3
Write \(3a\) outside the bracket.
\(3a(2 + 3b)\)

Checking your answer

Because factorising is the reverse of expanding, you can always check by expanding your answer. If it returns the original expression, you are right. This self check earns the Communicating marks under Criterion C.

ExpressionHCFFactorised
\(4x + 8\)\(4\)\(4(x + 2)\)
\(10y - 15\)\(5\)\(5(2y - 3)\)
\(12p + 8pq\)\(4p\)\(4p(3 + 2q)\)

Check yourself

1. Factorise \(8x + 12\). +

The HCF of \(8\) and \(12\) is \(4\). Dividing: \(8x \div 4 = 2x\), \(12 \div 4 = 3\). Answer: \(4(2x + 3)\).

2. Factorise \(15y - 20\). +

The HCF of \(15\) and \(20\) is \(5\). Dividing: \(15y \div 5 = 3y\), \(20 \div 5 = 4\). Answer: \(5(3y - 4)\).

3. Factorise \(6a + 9ab\). +

Both terms share \(3\) and \(a\), so the HCF is \(3a\). Dividing: \(6a \div 3a = 2\), \(9ab \div 3a = 3b\). Answer: \(3a(2 + 3b)\).

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