Simple Factorising (Factorising Expressions)
Factorising is expanding in reverse: the highest common factor (HCF) is taken outside a bracket, as in \(10p^2 - 15p = 5p(2p - 3)\). This NSW Year 9 Advanced maths subtopic extends Stage 4 factorising to negative and binomial common factors (NESA outcome MA5‑ALG‑P‑01).
You will factorise expressions with several pronumerals and powers, take out a negative HCF, and check an answer by substituting values. You will take out a common bracket, as in \(6(b - 2) - b(b - 2)\), factorise four-term expressions by grouping in pairs, factorise \(4x(x - 3)^2 - 8x^2(x - 3)\) fully, and use factors to explain why \(n^2 + n\) is always even. It is part of Year 9 Advanced Algebra: Expanding and Factorising.
Theory
To factorise, take out the highest common factor (HCF) of every term and write what is left inside a bracket. This Year 9 Advanced maths page covers HCFs with pronumerals and indices, a negative HCF, a bracket as the common factor and grouping in pairs for four-term expressions.
Factorising is the reverse of expanding. It writes a sum of terms as a product: \(10p^2 - 15p = 5p(2p - 3)\).
The highest common factor (HCF) of the terms is the HCF of the numbers times the lowest power of each pronumeral that appears in every term.
An expression is factorised fully when the bracket has no common factor left. \(8x^2 + 12x = 2x(4x + 6)\) is not fully factorised; \(4x(2x + 3)\) is.
When the first term is negative, it is usual to take out a negative HCF. Every sign inside the bracket then changes: \(-14y - 21y^2 = -7y(2 + 3y)\).
The common factor can be a whole bracket: \(a(k + 9) + 5(k + 9)\) \(= (k + 9)(a + 5)\).
The distributive law, read from right to left:
Grouping in pairs:
How to factorise by the HCF
- Find the HCF of the numbers, then the lowest power of each pronumeral in every term. Make it negative if the first term is negative.
- Divide each term by the HCF. The results go inside the bracket, one for each term.
- Check by expanding, or by substituting values. The bracket should have no common factor left.
Four terms: grouping in pairs
- Take out any HCF of all four terms first.
- Pair the terms (reorder if needed) so each pair has a common factor.
- Factorise each pair so the two brackets are identical. Take out a negative if the third term is negative.
- Take out the common bracket.
Using factors to explain
- Even numbers: \(n^2 - n = n(n - 1)\) is a product of two consecutive whole numbers, one of which is even, so \(n^2 - n\) is always even.
The HCF of \(18\), \(24\) and \(6\) is \(6\). Every term has at least \(a^2\) and at least \(b\). So the HCF is \(6a^2 b\).
| \(18a^3 b^2 \div 6a^2 b\) | \(=\) | \(3ab\) |
| \(-24a^2 b^3 \div 6a^2 b\) | \(=\) | \(-4b^2\) |
| \(6a^2 b \div 6a^2 b\) | \(=\) | \(1\) |
| \(18a^3 b^2 - 24a^2 b^3 + 6a^2 b\) | \(\) | \(\) |
| \(\) | \(=\) | \(\boldsymbol{6a^2 b(3ab - 4b^2 + 1)}\) |
When a term equals the HCF, a \(1\) is left behind in the bracket. Three terms in, three terms out.
(i) \(3x(y - 5) - (y - 5)\)
(ii) \(4(m - 3) + m(3 - m)\)
(i) Both terms contain the bracket \((y - 5)\). The second term is \(-1 \times (y - 5)\).
| \(3x(y - 5) - 1(y - 5)\) | \(=\) | \(\boldsymbol{(y - 5)(3x - 1)}\) |
(ii) The brackets are back to front. Use \(3 - m = -(m - 3)\) to make them match.
| \(4(m - 3) + m(3 - m)\) | \(\) | \(\) |
| \(\) | \(=\) | \(4(m - 3) - m(m - 3)\) |
| \(\) | \(=\) | \(\boldsymbol{(m - 3)(4 - m)}\) |
Check (ii) with \(m = 5\): \(4(2) + 5(-2) = -2\) and \((2)(-1) = -2\).
Four terms with no common factor: reorder so each pair shares a factor, then factorise each pair.
| \(ab + 12 - 4a - 3b\) | \(=\) | \(ab - 4a - 3b + 12\) |
| \(\) | \(=\) | \(a(b - 4) - 3(b - 4)\) |
| \(\) | \(=\) | \(\boldsymbol{(b - 4)(a - 3)}\) |
Take out \(-3\), not \(3\), from the second pair, so the brackets match: \(-3b + 12 = -3(b - 4)\). Expand to check: \(ab - 3b - 4a + 12\).
Both terms contain \(3\), \(x\) and \((x + 2)\). The HCF is \(3x(x + 2)\).
| \(\) | \(\) | \(3x(x + 2)\big[2(x + 2) - 3x\big]\) |
| \(\) | \(=\) | \(3x(x + 2)(2x + 4 - 3x)\) |
| \(\) | \(=\) | \(\boldsymbol{3x(x + 2)(4 - x)}\) |
Simplify inside the square bracket before writing the answer. Check with \(x = 1\): \(6(9) - 9(3) = 27\) and \(3(3)(3) = 27\).
Common pitfalls
Frequently asked questions
How do you factorise an expression?
Find the highest common factor of all the terms, write it in front of a bracket, and divide each term by it to get what goes inside. For example \(15x + 20 = 5(3x + 4)\).
How do you find the HCF of algebraic terms?
Take the HCF of the numbers, then each pronumeral that appears in every term, using its lowest power. The HCF of \(12a^3 b\) and \(8a^2 b^2\) is \(4a^2 b\).
What does factorise fully mean?
It means take out the highest common factor so that nothing common is left inside the bracket. 6x + 12 = 2(3x + 6) is not fully factorised because 3 is still common; 6(x + 2) is.
How do you factorise by grouping?
Split four terms into two pairs that each have a common factor, factorise each pair so the same bracket appears twice, then take that bracket out. For example xy + 2x + 3y + 6 = x(y + 2) + 3(y + 2) = (y + 2)(x + 3).
When do you take out a negative common factor?
Usually when the first term is negative. Taking out the negative makes the first term inside the bracket positive, and every sign inside changes.