Factorising Expressions
Every question with a fully worked solution.
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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, three terms and a negative HCF, how to take out a common bracket, and how to find a side of a rectangle from its area.
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.
Factorising is the reverse of expanding, so every factorisation can be checked by expanding it again.
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 take out a common bracket
- Look for a bracket that appears in every term, such as \((x + 1)\) in \(a(x + 1) + 3(x + 1)\).
- Treat the bracket like a single pronumeral and take it out. What is left of each term goes in the second bracket.
- Reverse a bracket if needed, using \((4 - x) = -(x - 4)\).
(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) \(8p^2q - 12pq^2 + 4pq\)
(ii) A rectangle has area \((12a^2 + 18ab)\) cm\(^2\) and width \(6a\) cm. Find an expression for its length.
(i) The HCF of \(8\), \(12\) and \(4\) is \(4\), and every term contains \(p\) and \(q\). The HCF is \(4pq\). The last term equals the HCF, so it leaves \(1\) in the bracket.
\(8p^2q - 12pq^2 + 4pq\)
| \(\) | \(=\) | \(\boldsymbol{4pq(2p - 3q + 1)}\) |
(ii) Factorise the area with the width outside the bracket. The other factor is the length.
| \(12a^2 + 18ab\) | \(=\) | \(6a(2a + 3b)\) |
The length is \(\boldsymbol{(2a + 3b)}\) cm. Check: \(6a \times 2a = 12a^2\) and \(6a \times 3b = 18ab\). ✓
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.
Can a bracket be a common factor?
Yes. In \(3x(x + 2) + 5(x + 2)\) the bracket \((x + 2)\) is in both terms, so the expression factorises to \((x + 2)(3x + 5)\).
How can I check a factorisation?
Expand your answer. You should get back the expression you started with.