To learn Algebraic Expansion and Factorisation quickly and correctly, think of these two skills as working in opposite directions. Expansion removes brackets by multiplying terms correctly; factorisation rebuilds a product by putting brackets back. When you understand the distributive law and can check one direction with the other, algebra becomes easier to read and mistakes become quicker to find. These eduKateSG Mathematics Quick Notes provide clear steps for common factors, simple quadratic expressions and reliable checking.
- Expand: multiply every term inside a bracket by the term outside.
- Combine like terms: add coefficients only when the variable parts match.
- Factorise: look first for a common factor shared by every term.
- For two brackets: multiply each term in one bracket by each term in the other.
- Recognise patterns: a² − b² = (a − b)(a + b).
- Check: expand your factorised answer to recover the original expression.
First, expand a single bracket using an everyday idea. Suppose you have 3(x + 4). This means three copies of (x + 4): (x + 4) + (x + 4) + (x + 4). There are three x terms and twelve units, so the result is 3x + 12. The compact rule is a(b + c) = ab + ac. Every term inside the bracket gets multiplied, including negative terms. For example, 2(x − 5) = 2x − 10. You can check with a simple value such as x = 6 before relying on a rule.
Second, factorise by reading the expanded expression backwards. In 6x + 9, both terms share the factor 3. Divide each term by 3 and write 3(2x + 3). Expand it again to check: 3 × 2x + 3 × 3 = 6x + 9. The shared factor must divide every term of the expression. If the expression is 8x² + 12x, both terms contain 4x, so it can be written 4x(2x + 3). Recognising a common factor first often makes longer factorisation questions much easier.
Third, move carefully between two brackets and a quadratic expression. Multiply (x + 2)(x + 3) by distributing each term: x × x = x², x × 3 = 3x, 2 × x = 2x and 2 × 3 = 6. Combine 3x + 2x to get x² + 5x + 6. To reverse this, look for two numbers whose product is 6 and whose sum is 5. The numbers are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3). Expansion confirms the factorisation.
Worked example: expand two brackets with a minus sign
Expand (x + 4)(x − 2). Multiply each term: x × x = x²; x × (−2) = −2x; 4 × x = 4x; and 4 × (−2) = −8. Combine the middle terms: −2x + 4x = 2x. The result is x² + 2x − 8. Check with x = 3: the original is (7)(1) = 7, while the expanded expression is 9 + 6 − 8 = 7. The check agrees.
Worked example: take out a common factor
Factorise 12x² − 18x. The coefficients 12 and 18 share the factor 6, and both terms contain x. Take out 6x: 12x² ÷ 6x = 2x, and −18x ÷ 6x = −3. Therefore 12x² − 18x = 6x(2x − 3). Expand to check: 6x × 2x − 6x × 3 = 12x² − 18x. The factorised result must reproduce both original terms, including their signs.
Worked example: difference of two squares
A useful pattern is a² − b² = (a − b)(a + b). Why does it work? Expanding the right side gives a² + ab − ab − b²; the middle terms cancel. For example, x² − 25 = (x − 5)(x + 5), because 25 is 5². This shortcut applies to a difference of squares, not an ordinary sum of squares. In introductory real-number factorisation, x² + 25 cannot be factorised into real linear factors.
A simple strategy for factorising expressions
- Read the whole expression and identify its terms.
- Check whether every term shares a number or variable factor.
- If it is a two-term difference of squares, test the (a − b)(a + b) identity.
- If it is x² + bx + c, look for two numbers whose sum is b and whose product is c.
- Write the product of factors clearly.
- Expand the answer to confirm that the original expression returns.
The difference between expanding and solving
The expression (x + 2)(x − 5) can be expanded to x² − 3x − 10, but that task does not give one numerical value for x. To solve an equation such as (x + 2)(x − 5) = 0, you would use the zero-product property and find x = −2 or x = 5. An expression and an equation are not the same task. Never invent an equals zero sign unless the question provides one.
Common algebra errors and how to repair them
- 2(x + 4) = 2x + 4: incorrect; both terms must be multiplied, giving 2x + 8.
- (x + 2)(x + 3) = x² + 6: incorrect; both middle products must be included.
- Combining x² and x as like terms: incorrect; they have different powers.
- Forgetting the sign of a negative product: positive × negative is negative.
- Taking a factor out of only one term: a common factor must divide every term.
- Not expanding to check: the reverse operation is the fastest error test.
Mini-test: do one of each
- Expand 3(x − 5).
- Factorise 8x + 12 using the greatest common positive integer factor.
- Expand (x + 4)(x − 2).
- Factorise x² + 7x + 12 into two linear brackets.
- Factorise x² − 16.
Answers with quick checks
1. 3x − 15. 2. 4(2x + 3). 3. x² + 2x − 8. 4. (x + 3)(x + 4), because 3 + 4 = 7 and 3 × 4 = 12. 5. (x − 4)(x + 4). Expand each factorised expression to confirm you get the original.
Ten minutes of useful practice
- Minutes 1–2: expand two single brackets, one with subtraction.
- Minutes 3–5: expand one product of two brackets and collect like terms.
- Minutes 6–7: factorise one expression with a common factor.
- Minute 8: try one difference-of-squares example.
- Minutes 9–10: expand your factorised answers to check them without looking back.
Learning both directions together is more effective than treating them as unrelated tricks. If you can explain why the expanded and factorised forms give the same value for every allowed x, you understand the algebra.
Keep learning: your next Quick Notes
Build this skill with Algebra · Linear Equations · Quadratic Equations · undefined. Pick a different question and explain why your method works without looking at the example.
For the wider Secondary Mathematics learning pathway, read Secondary 1 Mathematics Tutor Clementi | Small Groups Tutorials. The original reference page remains unchanged.