Factorising Algebraic Expressions
Every question with a fully worked solution.
- Factorising Algebraic Expressions - Video - Factorisation Watch
Theory
Factorising is expanding in reverse: find what every term has in common and put it outside a bracket. This Year 9 maths page shows how to find the highest common factor, factorise expressions with two or three terms, take out a negative HCF or a whole bracket, and check the answer by expanding.
To factorise an expression means to write it as a product. It is the reverse of expanding: \(3(x + 4)\) expands to \(3x + 12\), so \(3x + 12\) factorises to \(3(x + 4)\).
The highest common factor (HCF) of the terms is the largest number AND the highest power of each pronumeral that divides into every term. A pronumeral missing from even one term is not part of the HCF.
An expression is fully factorised when nothing more can be taken out of the bracket.
The HCF can be negative, or it can be a whole bracket, such as \((y + 4)\) in \(k(y + 4) + 3(y + 4)\).
Factorising takes the common factor out:
Taking out a negative flips every sign inside:
| Expression | Number part | Letter part | HCF |
|---|---|---|---|
| \(10a + 15\) | \(5\) | none (no \(a\) in \(15\)) | \(5\) |
| \(7x^2 + 7x\) | \(7\) | \(x\) | \(7x\) |
| \(9mn - 6m^2\) | \(3\) | \(m\) | \(3m\) |
How to factorise by taking out the HCF
- Find the largest number that divides every coefficient.
- Find the pronumerals in every term, with the lowest power of each. Together with the number, this is the HCF. If the first term is negative, use a negative HCF.
- Divide each term by the HCF and write the results inside a bracket, with the HCF outside.
- Check that nothing more is common inside the bracket, then expand to check the answer.
(i) \(15x + 25\) and
(ii) \(4p^2 + 10p - 6pq\).
(i) The HCF of \(15\) and \(25\) is \(5\). There is no \(x\) in \(25\), so the HCF is just \(5\):
| \(15x + 25\) | \(=\) | \(\boldsymbol{5(3x + 5)}\) |
(ii) The numbers \(4\), \(10\) and \(6\) share \(2\), and every term has a \(p\). The HCF is \(2p\). Divide each term by it:
| \(4p^2 \div 2p\) | \(=\) | \(2p\) |
| \(10p \div 2p\) | \(=\) | \(5\) |
| \(-6pq \div 2p\) | \(=\) | \(-3q\) |
| \(4p^2 + 10p - 6pq\) | \(=\) | \(\boldsymbol{2p(2p + 5 - 3q)}\) |
Check by expanding: \(2p \times 2p = 4p^2\), \(2p \times 5 = 10p\), \(2p \times (-3q) = -6pq\).
Build the HCF in two pieces: the number part and the letter part.
| \(\text{numbers}\) | \(:\) | \(\text{HCF of } 8 \text{ and } 20 = 4\) |
| \(\text{letters}\) | \(:\) | \(ab \text{ is in both}\) |
| \(\text{HCF}\) | \(=\) | \(4ab\) |
| \(8a^2b \div 4ab\) | \(=\) | \(2a\) |
| \(20ab^2 \div 4ab\) | \(=\) | \(5b\) |
| \(8a^2b - 20ab^2\) | \(=\) | \(\boldsymbol{4ab(2a - 5b)}\) |
Fully factorised: \(2a\) and \(5b\) have nothing more in common. Taking out only \(2ab\) would leave \(4a - 10b\), which still has \(2\) in common.
The HCF of \(6\) and \(14\) is \(2\). The first term is negative, so take out \(-2\). Dividing by a negative flips each sign:
| \(-6m \div (-2)\) | \(=\) | \(3m\) |
| \(-14 \div (-2)\) | \(=\) | \(7\) |
| \(-6m - 14\) | \(=\) | \(\boldsymbol{-2(3m + 7)}\) |
Check: \(-2 \times 3m = -6m\) and \(-2 \times 7 = -14\).
(i) \(k(y + 4) + 3(y + 4)\) and
(ii) \(t(t - 5) - (t - 5)\).
(i) Both terms contain the whole bracket \((y + 4)\), so it comes out. What is left is \(k\) and \(3\):
| \(k(y + 4) + 3(y + 4)\) | \(=\) | \(\boldsymbol{(y + 4)(k + 3)}\) |
(ii) The second term is \(-1\) lots of \((t - 5)\). Write the hidden \(1\) in:
| \(t(t - 5) - 1(t - 5)\) | \(=\) | \(\boldsymbol{(t - 5)(t - 1)}\) |
Without the \(1\), the second bracket would be missing a term.
Common pitfalls
Frequently asked questions
How do you factorise an expression?
Find the highest common factor of all the terms, write it outside a bracket, and write what is left of each term inside the bracket.
How do you find the HCF of algebraic terms?
Find the largest number that divides all the coefficients, then include each pronumeral that appears in every term, with its lowest power.
What does factorise fully mean?
It means taking out the highest common factor, so that the terms left inside the bracket have no common factor other than \(1\).
How do you check a factorisation?
Expand your answer. If it gives back the original expression, the factorisation is correct.
How do you factorise with a negative HCF?
Take out the negative number as part of the HCF. Every sign inside the bracket changes. For example \(-6m - 14 = -2(3m + 7)\).