Factorising Expressions
Theory
Factorising is the reverse of the distributive law: \(ab + ac = a(b + c)\). Take the highest common factor (HCF) of the terms outside a bracket: \(12x + 3xy = 3x \times 4 + 3x \times y = 3x(4 + y)\). This Year 8 maths page shows how.
Expanding turns \(4(2y + 5)\) into \(8y + 20\). Factorising turns \(8y + 20\) back into \(4(2y + 5)\).
Factors of a term. The factors of \(10n\) are \(1, 2, 5, 10, n, 2n, 5n\) and \(10n\).
The HCF of the terms is the biggest number that divides every coefficient, times every pronumeral in every term, each to its lowest power. For \(15x^2y\) and \(3xy^2\) it is \(3xy\).
Three steps. Find the HCF. Divide each term by it. Write the HCF outside the bracket and the answers inside. A term divided by itself leaves \(1\): \(9w + 9wz = 9w(1 + z)\).
Check by expanding: you must get the expression you started with. A bracket can be the common factor: \(a(x + y) + b(x + y) = (x + y)(a + b)\).
Factorising reverses expanding:
How to factorise
- Find the HCF of the numbers and of the pronumerals (lowest powers).
- Divide each term by the HCF.
- Write HCF(answers) and check by expanding.
| \(12x + 3xy\) | \(=\) | \(3x \times 4 + 3x \times y\) |
| \(\) | \(=\) | \(3x(4 + y)\) |
\(3x(4 + y)\)
| \(\text{HCF}\) | \(=\) | \(3xy\) |
| \(15x^2y - 3xy^2\) | \(=\) | \(3xy(5x - y)\) |
\(3xy(5x - y)\): \(15x^2y \div 3xy = 5x\) and \(3xy^2 \div 3xy = y\).
(i) \(6a + 15\)
(ii) \(10k - 4\)
| \(6a + 15\) | \(=\) | \(3(2a + 5)\) |
| \(10k - 4\) | \(=\) | \(2(5k - 2)\) |
(i) \(3(2a + 5)\)
(ii) \(2(5k - 2)\)
(i) \(n^2 + 7n\)
(ii) \(12ab - 20b\)
| \(n^2 + 7n\) | \(=\) | \(n(n + 7)\) |
| \(12ab - 20b\) | \(=\) | \(4b(3a - 5)\) |
(i) \(n(n + 7)\). Check with \(n = 2\): \(4 + 14 = 18 = 2 \times 9\).
(ii) \(4b(3a - 5)\)
| \(\text{HCF}\) | \(=\) | \(5p\) |
| \(\) | \(=\) | \(5p(2p - 3q + 1)\) |
\(5p(2p - 3q + 1)\): the last term gives \(5p \div 5p = 1\).
(ii) Factorise \(8(t - 1) + t(t - 1)\).
| \(\text{HCF}\) | \(=\) | \(7c^2d\) |
| \(14c^3d - 21c^2d^2\) | \(=\) | \(7c^2d(2c - 3d)\) |
| \(8(t - 1) + t(t - 1)\) | \(=\) | \((t - 1)(8 + t)\) |
(i) \(7c^2d(2c - 3d)\)
(ii) \((t - 1)(8 + t)\)
Common pitfalls
Frequently asked questions
What does factorise mean?
Write an expression as a product, with the HCF outside a bracket: \(6x + 9 = 3(2x + 3)\).
How do you find the HCF of algebraic terms?
Take the HCF of the numbers and every pronumeral common to all terms, at its lowest power.
How do you check a factorisation?
Expand the bracket; you should get the original expression.
Why is there a 1 in some brackets?
A term divided by itself is \(1\): \(4a + 4ab = 4a(1 + b)\).
Can a bracket be a common factor?
Yes: \(x(y + 2) + 3(y + 2) = (y + 2)(x + 3)\).