What you'll learn
Factorisation forms a critical component of CXC CSEC Mathematics Paper 2, appearing regularly in questions worth 4-8 marks. This revision guide covers the three essential factorisation techniques tested at CSEC level: extracting common factors, recognising and factorising the difference of two squares, and factorising quadratic trinomials. These skills serve as building blocks for solving equations, simplifying algebraic fractions, and tackling real-world problems involving area, profit, and rates.
Key terms and definitions
Factorisation — the process of writing an algebraic expression as a product of its factors, essentially the reverse of expansion.
Common factor — a term (number, variable, or both) that divides exactly into every term of an algebraic expression.
Difference of two squares — an expression in the form a² − b², which always factorises to (a + b)(a − b).
Quadratic trinomial — a three-term polynomial of degree 2, typically written as ax² + bx + c, where a, b, and c are constants.
Prime expression — an algebraic expression that cannot be factorised further using integers or simple algebraic terms.
Leading coefficient — the coefficient of the highest power term in a polynomial; in 3x² + 5x − 2, the leading coefficient is 3.
Binomial — an algebraic expression containing exactly two terms, such as (x + 5) or (2a − 3b).
Perfect square — a number or algebraic term that is the square of an integer or algebraic expression, such as 4, 9, x², or 25y².
Core concepts
Factorising by extracting common factors
The first step in any factorisation problem involves identifying the highest common factor (HCF) of all terms. This technique appears in approximately 70% of CSEC factorisation questions, often as a preliminary step before applying other methods.
Process for extracting common factors:
- Identify the HCF of all numerical coefficients
- Identify the lowest power of each variable that appears in every term
- Write the HCF outside brackets
- Divide each original term by the HCF to find what goes inside the brackets
- Check your answer by expanding
Examples of common factor extraction:
- 6x + 9 = 3(2x + 3)
- 12a²b − 8ab² = 4ab(3a − 2b)
- 5x²y + 15xy² − 10xy = 5xy(x + 3y − 2)
Caribbean context: When calculating the total cost of mangoes at x dollars per kilogram and pineapples at y dollars per kilogram, the expression 3x + 3y factorises to 3(x + y), showing that buying 3 kg of each fruit costs 3 times the combined unit price.
Grouping method:
When expressions have four terms with no overall common factor, group terms in pairs:
- ax + ay + bx + by = a(x + y) + b(x + y) = (x + y)(a + b)
- 3m − 6n + mp − 2np = 3(m − 2n) + p(m − 2n) = (m − 2n)(3 + p)
Difference of two squares
This special pattern appears frequently in CXC CSEC papers and must be recognised instantly. The formula a² − b² = (a + b)(a − b) applies whenever you have one perfect square subtracted from another.
Recognition checklist:
- Exactly two terms separated by a minus sign
- Both terms are perfect squares (coefficients are 1, 4, 9, 16, 25... and variables have even powers)
- No addition, only subtraction
Common CSEC examples:
- x² − 25 = (x + 5)(x − 5)
- 9a² − 16b² = (3a + 4b)(3a − 4b)
- 49 − y² = (7 + y)(7 − y)
- 4x² − 1 = (2x + 1)(2x − 1)
- x²y² − 36 = (xy + 6)(xy − 6)
Two-stage factorisation:
Sometimes you must extract a common factor first, then apply the difference of two squares:
- 2x² − 50 = 2(x² − 25) = 2(x + 5)(x − 5)
- 3a²b − 27b = 3b(a² − 9) = 3b(a + 3)(a − 3)
Caribbean context: The area of a square cricket pitch of side x metres minus the area of a square pavilion of side 5 metres gives x² − 25, which factorises to (x + 5)(x − 5) square metres.
Factorising quadratic trinomials (a = 1)
When the coefficient of x² is 1, the trinomial x² + bx + c factorises to (x + p)(x + q), where p and q are numbers satisfying two conditions simultaneously:
The two-number rule:
- p + q = b (the sum equals the coefficient of x)
- p × q = c (the product equals the constant term)
Step-by-step method:
- Write x² + bx + c = (x )(x )
- Find two numbers that multiply to give c and add to give b
- Place these numbers in the brackets: (x + p)(x + q)
- Check by expanding
Sign patterns to memorise:
| Signs in trinomial | Signs in factors | Example |
|---|---|---|
| x² + bx + c (both +) | both + | x² + 7x + 12 = (x + 3)(x + 4) |
| x² − bx + c (middle −, end +) | both − | x² − 7x + 12 = (x − 3)(x − 4) |
| x² + bx − c (end −) | one + one − (larger absolute value matches b's sign) | x² + x − 12 = (x + 4)(x − 3) |
| x² − bx − c (end −) | one + one − | x² − x − 12 = (x − 4)(x + 3) |
CSEC-level examples:
- x² + 8x + 15 = (x + 3)(x + 5) because 3 + 5 = 8 and 3 × 5 = 15
- x² − 9x + 20 = (x − 4)(x − 5) because −4 + (−5) = −9 and −4 × (−5) = 20
- x² + 2x − 35 = (x + 7)(x − 5) because 7 + (−5) = 2 and 7 × (−5) = −35
Factorising quadratic trinomials (a ≠ 1)
When the coefficient of x² is not 1, two methods appear in CSEC exams: the inspection method (trial and error) and the grouping method (splitting the middle term).
Grouping method (most reliable for exams):
For ax² + bx + c:
- Multiply a × c to find the target product
- Find two numbers that multiply to give ac and add to give b
- Rewrite bx as the sum of these two numbers multiplied by x
- Group in pairs and factorise each pair
- Extract the common binomial factor
Worked through: 2x² + 7x + 3
- a × c = 2 × 3 = 6
- Need two numbers: product = 6, sum = 7 → numbers are 6 and 1
- Rewrite: 2x² + 6x + 1x + 3
- Group: 2x(x + 3) + 1(x + 3)
- Factorise: (x + 3)(2x + 1)
Inspection method:
For 3x² + 10x + 8:
- Factors of 3x²: (3x )(x )
- Factors of 8: could be 1 and 8, or 2 and 4
- Test combinations: (3x + 2)(x + 4) expands to 3x² + 14x + 8 ✗
- Try: (3x + 4)(x + 2) expands to 3x² + 10x + 8 ✓
Common CSEC examples:
- 2x² + 5x + 2 = (2x + 1)(x + 2)
- 3x² − 10x + 8 = (3x − 4)(x − 2)
- 6x² + x − 2 = (2x − 1)(3x + 2)
Caribbean context: The profit P in dollars from selling n jerk chicken meals is sometimes modelled as P = 2n² + 9n − 5, which factorises to (2n − 1)(n + 5), useful for finding break-even points.
Complete factorisation strategy
CXC examiners test your ability to apply multiple techniques in sequence. Always work through this checklist:
- Common factor first — extract any HCF from all terms
- Count terms — two terms? Check for difference of two squares. Three terms? Quadratic trinomial. Four terms? Try grouping.
- Check each factor — can any factor be factorised further?
- Verify by expanding — multiply out your answer to confirm it matches the original
Multi-step examples:
- 4x² − 36 = 4(x² − 9) = 4(x + 3)(x − 3)
- 3x² + 12x + 9 = 3(x² + 4x + 3) = 3(x + 1)(x + 3)
- 2x³ − 8x = 2x(x² − 4) = 2x(x + 2)(x − 2)
Worked examples
Example 1: Multi-stage factorisation (typical 4-mark question)
Factorise completely: 5x² − 45
Solution:
Step 1: Extract the common factor 5x² − 45 = 5(x² − 9) [1 mark]
Step 2: Recognise difference of two squares x² − 9 = x² − 3²
Step 3: Apply the formula a² − b² = (a + b)(a − b) 5(x² − 9) = 5(x + 3)(x − 3) [1 mark for each factor]
Answer: 5(x + 3)(x − 3) [1 mark for complete factorisation]
Example 2: Quadratic trinomial (typical 5-mark question)
A rectangular garden in Barbados has length (x + 7) metres and width (x + 2) metres. The area is 60 m².
(a) Write an equation for the area. [2 marks] (b) Simplify the equation to form a quadratic equation equal to zero. [2 marks] (c) Factorise your quadratic equation. [3 marks]
Solution:
(a) Area = length × width (x + 7)(x + 2) = 60 [1 mark for expression, 1 mark for equation]
(b) Expand: x² + 2x + 7x + 14 = 60 Simplify: x² + 9x + 14 = 60 Subtract 60: x² + 9x − 46 = 0 [1 mark for expansion, 1 mark for standard form]
(c) Need two numbers: product = −46, sum = 9 These numbers do not factorise neatly with integers, so check the question or use the quadratic formula. [In actual exam, the numbers would be chosen to factorise.]
Corrected version for factorisation: If the area were 60 m² and simplified to x² + 9x − 36 = 0: Numbers: 12 and −3 (since 12 × −3 = −36 and 12 + (−3) = 9) x² + 9x − 36 = (x + 12)(x − 3) [2 marks for correct factors, 1 mark for complete answer]
Example 3: Grouping method (typical 6-mark question)
Factorise: 6x² + 11x + 4
Solution:
Step 1: Identify a = 6, b = 11, c = 4 Calculate a × c = 6 × 4 = 24 [1 mark]
Step 2: Find two numbers that multiply to 24 and add to 11 Numbers are 8 and 3 (8 × 3 = 24, 8 + 3 = 11) [1 mark]
Step 3: Rewrite the middle term 6x² + 11x + 4 = 6x² + 8x + 3x + 4 [1 mark]
Step 4: Group in pairs = (6x² + 8x) + (3x + 4) [1 mark]
Step 5: Factorise each pair = 2x(3x + 4) + 1(3x + 4) [1 mark]
Step 6: Extract common factor = (3x + 4)(2x + 1) [1 mark]
Answer: (3x + 4)(2x + 1)
Check: (3x + 4)(2x + 1) = 6x² + 3x + 8x + 4 = 6x² + 11x + 4 ✓
Common mistakes and how to avoid them
• Mistake: Forgetting to extract the common factor first. Students jump straight to factorising 2x² − 18 as a difference of squares, writing (2x + 6)(2x − 6), which is incorrect. Correction: Always check for a common factor first: 2x² − 18 = 2(x² − 9) = 2(x + 3)(x − 3).
• Mistake: Incorrect signs in difference of two squares. Writing x² − 25 = (x − 5)(x − 5) instead of (x + 5)(x − 5). Correction: The formula a² − b² = (a + b)(a − b) requires one plus and one minus sign. Check by expanding: (x + 5)(x − 5) = x² − 25 ✓, but (x − 5)(x − 5) = x² − 10x + 25 ✗.
• Mistake: Confusing sum of squares with difference of squares. Attempting to factorise x² + 25 using real numbers. Correction: The sum of two squares (a² + b²) cannot be factorised using real numbers at CSEC level. Only the difference of two squares (a² − b²) factorises.
• Mistake: Wrong sign combinations in quadratic trinomials. For x² − 5x − 6, writing (x − 2)(x − 3) because "there's a minus sign." Correction: When the constant term is negative, the factors have opposite signs. Need two numbers: product = −6, sum = −5. Numbers are −6 and +1, giving (x − 6)(x + 1).
• Mistake: Not checking if factors can be factorised further. Stopping at 3x² − 12 = (3x + 6)(x − 2) without factorising the common factor from 3x + 6. Correction: After factorising, check each factor. Here, extract 3 from the first factor: 3x² − 12 = 3(x² − 4) = 3(x + 2)(x − 2).
• Mistake: Arithmetic errors when splitting the middle term. In 3x² + 10x + 8, calculating 3 × 8 = 24 but then choosing numbers that add to 24 instead of 10. Correction: Find two numbers where product = ac = 24 AND sum = b = 10. The numbers are 6 and 4 (not just any factors of 24).
Exam technique for factorisation in CXC CSEC Mathematics
• Recognise command words precisely. "Factorise" means write as a product of factors—marks are lost for simply rearranging or expanding. "Factorise completely" or "Factorise fully" means continue until no factor can be broken down further. Questions stating "Hence solve..." require you to use your factorised form, so accuracy in factorisation secures marks in subsequent parts.
• Show all working systematically. CXC mark schemes allocate method marks for intermediate steps. When factorising 6x² + 13x + 6, showing "ac = 36, need 9 and 4" then "6x² + 9x + 4x + 6" earns partial credit even if you make an error later. Jumping to an answer risks earning zero if incorrect.
• Check your answers within the time available. Expanding your factorised answer takes 15-20 seconds but prevents losing 3-4 marks. In Paper 2 Section II questions, factorisation often appears as part (a) worth 3 marks, with parts (b) and (c) building on it—a factorisation error cascades through the question.
• Allocate time according to marks. A 3-mark factorisation should take approximately 2-3 minutes. If you spend 6 minutes trying different factor combinations, move on and return later. The difference of two squares typically earns 2 marks and should be completed in under one minute once recognised.
Quick revision summary
Factorisation reverses expansion by writing expressions as products. Always extract common factors first from all terms. Recognise the difference of two squares pattern a² − b² = (a + b)(a − b) instantly—check for two perfect squares separated by subtraction. For quadratic trinomials x² + bx + c, find two numbers that multiply to c and add to b. When the leading coefficient a ≠ 1, use the grouping method: multiply ac, split the middle term, group pairs, and extract the common binomial. Apply multiple techniques in sequence for complete factorisation and always verify by expanding.