Factorising Expressions
Factorising Expressions in NSW Year 10 Core maths is the reverse of expanding (NESA outcome MA5‑ALG‑P‑01). The first step is always to take out the highest common factor, including pronumerals and, where helpful, a negative sign, so \(6x^2 - 18x = 6x(x - 3)\).
You will factorise fully using the highest common factor, take out a common binomial factor such as \((q + 3)\), and factorise four-term expressions by grouping them in pairs (MA5‑ALG‑P‑02). You will also notice when \((4 - x)\) can be rewritten as \(-(x - 4)\), and factorise the area of a garden bed to find an expression for its length. It is part of Year 10 Core Quadratics.
Every question with a fully worked solution.
- Factorising Expressions - Video - Factorising - Highest Common Factor Types Watch
Theory
Factorising is the reverse of expanding: it writes an expression as a product. This Year 10 maths page shows how to take out the highest common factor (HCF), including powers and a negative HCF, how to take out a common bracket, and how to factorise four terms by grouping in pairs. This content comes up if you're heading for the Advanced path in Year 11.
A factor is something that divides exactly into a term. To factorise an expression is to write it as a product of factors, using brackets.
The highest common factor (HCF) of two or more terms is the largest number and the highest power of each pronumeral that divides into every term. For \(15a^2b\) and \(10ab^3\) the HCF is \(5ab\).
To factorise fully means that no common factor is left inside the bracket. \(4x^2 + 8x = 2x(2x + 4)\) is not fully factorised, because \(2\) still divides into \(2x + 4\). The full answer is \(4x(x + 2)\).
A common bracket can be a factor too. In \(a(x + 1) + 3(x + 1)\), the bracket \((x + 1)\) is in both terms.
Grouping in pairs is a method for four terms: factorise each pair, then take out the common bracket.
Factorising uses the distributive law backwards:
How to factorise using the HCF
- Find the HCF of the numbers.
- Find the lowest power of each pronumeral that appears in every term. A pronumeral missing from one term is not in the HCF.
- Write the HCF outside a bracket. Divide each term by the HCF to get the terms inside. If a term equals the HCF, write \(1\) inside.
- Use a negative HCF when asked, or when the first term is negative. Every sign inside the bracket changes.
- Check by expanding, and make sure nothing more can be taken out.
How to factorise four terms by grouping
- Group the terms in two pairs. Take the HCF out of each pair. If the third term is negative, take out a negative factor.
- Compare the brackets. They must be identical; if not, try reordering the terms.
- Take out the common bracket, then check by expanding.
(i) \(18y^2 - 12y\)
(ii) \(12m^3n^2 + 20m^2n^3\)
Find the HCF of the numbers, then the lowest power of each pronumeral that is in every term.
(i) The HCF of \(18\) and \(12\) is \(6\), and both terms contain \(y\). The HCF is \(6y\).
| \(18y^2 - 12y\) | \(=\) | \(6y \times 3y - 6y \times 2\) |
| \(\) | \(=\) | \(\boldsymbol{6y(3y - 2)}\) |
(ii) The HCF of \(12\) and \(20\) is \(4\). The lowest powers are \(m^2\) and \(n^2\). The HCF is \(4m^2n^2\).
| \(12m^3n^2 + 20m^2n^3\) | \(=\) | \(4m^2n^2 \times 3m\) \({} + 4m^2n^2 \times 5n\) |
| \(\) | \(=\) | \(\boldsymbol{4m^2n^2(3m + 5n)}\) |
Check by expanding: \(4m^2n^2 \times 3m = 12m^3n^2\). ✓
(i) \(-10x^2 - 4x\), taking out a negative HCF
(ii) \(6a^3 - 9a^2 + 15a\)
(i) The HCF is \(-2x\). Dividing by a negative changes every sign inside the bracket.
| \(-10x^2 \div (-2x)\) | \(=\) | \(5x\) |
| \(-4x \div (-2x)\) | \(=\) | \(2\) |
| \(-10x^2 - 4x\) | \(=\) | \(\boldsymbol{-2x(5x + 2)}\) |
(ii) The HCF of \(6\), \(9\) and \(15\) is \(3\), and every term contains \(a\). The HCF is \(3a\). The bracket keeps three terms.
| \(6a^3 - 9a^2 + 15a\) | \(=\) | \(\boldsymbol{3a(2a^2 - 3a + 5)}\) |
(i) \(5y(y - 3) - 2(y - 3)\)
(ii) \(x(x - 6) + 4(6 - x)\)
(i) Both terms contain the bracket \((y - 3)\). Take it out; what is left goes in the second bracket.
| \(5y(y - 3) - 2(y - 3)\) | \(=\) | \(\boldsymbol{(y - 3)(5y - 2)}\) |
(ii) \((6 - x)\) is the negative of \((x - 6)\), so \(+4(6 - x) = -4(x - 6)\).
| \(x(x - 6) + 4(6 - x)\) | \(=\) | \(x(x - 6) - 4(x - 6)\) |
| \(\) | \(=\) | \(\boldsymbol{(x - 6)(x - 4)}\) |
(i) \(xy - 5x + 3y - 15\)
(ii) \(2ab - 6a - b + 3\)
(i) Take the HCF out of each pair.
| \(xy - 5x + 3y - 15\) | \(=\) | \(x(y - 5) + 3(y - 5)\) |
| \(\) | \(=\) | \(\boldsymbol{(y - 5)(x + 3)}\) |
(ii) The second pair starts with a minus, so take out \(-1\). Then the brackets match.
| \(2ab - 6a - b + 3\) | \(=\) | \(2a(b - 3) - 1(b - 3)\) |
| \(\) | \(=\) | \(\boldsymbol{(b - 3)(2a - 1)}\) |
Check: \((b - 3)(2a - 1) = 2ab - b - 6a + 3\). ✓
Common pitfalls
Frequently asked questions
What does factorise fully mean?
It means taking out the highest common factor, so that no number or pronumeral can still be taken out of the bracket.
How do you find the HCF of algebraic terms?
Find the HCF of the numbers, then take the lowest power of each pronumeral that is in every term. For \(15a^2b\) and \(10ab^3\) the HCF is \(5ab\).
When do you take out a negative HCF?
When a question asks for it, or when the first term is negative. Dividing by a negative changes the sign of every term inside the bracket.
How do you factorise four terms?
Group them in pairs and take the HCF out of each pair. If the two brackets are the same, take that bracket out as a common factor.
How can I check a factorisation?
Expand your answer. You should get back the expression you started with.