Secondary 4 G2 Mathematics Tutor Bukit Timah | Quadratic Factorisation and Algebraic Identities. Did you know? The expression x² + 7x + 12 is a small puzzle with two hidden numbers. They add to 7 and multiply to 12. Find 3 and 4, and the expression becomes (x + 3)(x + 4). That is not a trick; it is multiplication read backwards.
For a Secondary 4 G2 Mathematics student in Bukit Timah, quadratic expressions become easier when expansion, factorisation and checking are treated as one connected idea. A tutor who only supplies two numbers to put in brackets may help with today’s worksheet, but the stronger goal is to recognise the structure independently in an unfamiliar question.
This guide focuses on the N5 Algebraic expressions and formulae topic in the official 2027 SEC G2 Mathematics K210 syllabus: expanding products, common factors, identities such as (a + b)² and a² − b², and factorising ax² + bx + c. The older 2026 graduating cohort should use its school-confirmed N(A)-Level Mathematics Syllabus A 4045 examination entry. SEAB’s 2026 directory and the 2027 G2 directory distinguish those years.
Quick answer: the expansion-and-factorisation routine
- Look for a common factor. Every term may contain the same number, letter or algebraic bracket.
- Identify a known pattern. Two squared terms separated by a minus sign may be a difference of squares; three terms may form a perfect square.
- For a quadratic trinomial, find compatible pairs. The numbers or factors must reproduce both the middle coefficient and the final constant.
- Expand your factorised answer. If the result is not the starting expression, fix the factors before moving on.
- Keep factorising distinct from solving. An expression has factors; an equation equal to zero may additionally have roots.
At a glance: three algebraic identities worth understanding
| Identity | Expansion | Why the middle term appears |
|---|---|---|
| Square of a sum | (a + b)² = a² + 2ab + b² | The product ab appears twice. |
| Square of a difference | (a − b)² = a² − 2ab + b² | Two cross-products each contribute −ab. |
| Difference of squares | (a + b)(a − b) = a² − b² | The +ab and −ab terms cancel. |
Did you know? (a + b)² is not a² + b². Imagine a square of side a + b. It contains two smaller squares and two rectangles of area ab. That picture explains the missing 2ab term more reliably than a chant.
Twenty original worked examples: understand first, calculate second
1. Two positive binomials
Expand (x + 4)(x + 3): x² + 3x + 4x + 12 = x² + 7x + 12. Verify by checking the four products.
2. One positive and one negative constant
Expand (x − 5)(x + 2): x² + 2x − 5x − 10 = x² − 3x − 10. The final constant is negative because the constants have opposite signs.
3. Coefficient larger than one
Expand (2x + 3)(x + 4): 2x² + 8x + 3x + 12 = 2x² + 11x + 12. Combine only like terms.
4. Square of a sum
Expand (x + 6)²: x² + 2(x)(6) + 36 = x² + 12x + 36. The cross-term is 12x, not 6x.
5. Square of a difference
(x − 4)² = x² − 8x + 16. Both the first and final terms are positive squares. The negative is in the middle term.
6. A squared coefficient
(2x + 3)² = (2x)² + 2(2x)(3) + 3² = 4x² + 12x + 9. Square the whole first term.
7. Difference of squares expanded
(x + 7)(x − 7) = x² − 49. The two cross-products cancel, so there is no x term.
8. Another difference of squares
(3x + 2)(3x − 2) = (3x)² − 2² = 9x² − 4. The first square contains 3² = 9.
9. Factor out a numerical common factor
6x + 9 = 3(2x + 3). Confirm by distributing 3 across both bracket terms.
10. Factor out a letter and number
8x² + 12x = 4x(2x + 3). The common factor is 4x because both terms contain at least one x.
11. Monic positive quadratic
Factorise x² + 7x + 12. Find two numbers adding to 7 and multiplying to 12: 3 and 4. So (x + 3)(x + 4).
12. A negative constant
Factorise x² − x − 20. Numbers −5 and 4 add to −1 and multiply to −20, giving (x − 5)(x + 4).
13. Two negative bracket terms
Factorise x² − 9x + 20. Numbers −5 and −4 have sum −9 and product 20. Answer: (x − 5)(x − 4).
14. Leading coefficient 2
Factorise 2x² + 7x + 3. Reconstruct (2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3.
15. Leading coefficient 3
Factorise 3x² + 11x + 6. Use (3x + 2)(x + 3). Expanding gives 3x² + 9x + 2x + 6.
16. Difference of two squares factored
4x² − 9 = (2x)² − 3² = (2x − 3)(2x + 3). Check by expansion.
17. Perfect-square trinomial
x² + 6x + 9 = (x + 3)². The constant is 3² and the middle term is 2 × 3 × x.
18. Grouping with two variables
ax + bx + kay + kby = x(a + b) + ky(a + b) = (a + b)(x + ky). This example illustrates grouping rather than trial-and-error.
19. Use factors to solve an equation
If x² − 5x + 6 = 0, then (x − 2)(x − 3) = 0; hence x = 2 or x = 3. You used factorisation to solve because the expression was set equal to zero.
20. Area difference as an algebraic identity
The area between a square of side x + 5 and an inner square of side x is (x + 5)² − x² = 10x + 25. The difference-of-squares identity also gives 5(2x + 5).
How to tell whether a student understands the factors
| First wrong line | Hidden gap | Tutor’s useful follow-up |
|---|---|---|
| (x + 4)² = x² + 16 | Cross-product missing | Draw the square or expand the two brackets separately. |
| x² + 7x + 12 = (x + 7)(x + 12) | Pair chosen by appearance | What are the sum and product of your chosen constants? |
| 2x² + 7x + 3 = (2x + 3)(x + 1) | Middle coefficient not checked | Expand the factors and compare the coefficient of x. |
| x² − 25 = (x − 5)² | Difference of squares confused with a squared difference | Expand (x − 5)² and notice the unwanted −10x. |
| Solves x² + 4x + 4 as though it equals zero | Expression and equation confused | Does the question ask to factorise, evaluate or solve? |
| Factors 6x² + 9x without common factor | Missed the simplest first step | What common factor divides both terms? |
A changed six-question test you can try without notes
- Expand (x − 2)(x + 7).
- Expand (3x − 2)².
- Factorise x² + 9x + 20.
- Factorise x² − 16.
- Factorise 2x² + 9x + 4.
- Solve x² − 7x + 12 = 0.
Answers: (1) x² + 5x − 14; (2) 9x² − 12x + 4; (3) (x + 4)(x + 5); (4) (x − 4)(x + 4); (5) (2x + 1)(x + 4); (6) x = 3 or 4. For the factorisations, expand again to verify that each expression is exactly reproduced.
A useful 90-minute three-student tutorial
| Time | Teaching move | What progress looks like |
|---|---|---|
| 0–10 min | Retrieve identities from a delayed prior lesson | Student reconstructs a² − b² without reading it. |
| 10–25 min | Individually expand three unfamiliar expressions | Tutor identifies the first incorrect cross-product. |
| 25–50 min | Compare factorisation methods and algebraic identities | Student explains the required sum and product. |
| 50–75 min | Each learner tackles six mixed changed questions | Correct method selection without a chapter heading. |
| 75–90 min | Exit check, first-error log and delayed retest | Independent factorisation of a new trinomial later in the week. |
The unchanged eduKateSG three-pupil Mathematics small-group reference describes premium 1.5-hour weekly tutorials near Sixth Avenue MRT, guided corrections and individual attention. The above is an illustrative Secondary 4 G2 lesson design, not a claim that a particular class or vacancy is currently available.
An easy-to-follow seven-day revision loop
- Day 1: expand positive and negative brackets, narrating the four products.
- Day 2: explore perfect squares and difference-of-squares identities.
- Day 3: factorise expressions with simple common factors.
- Day 4: factorise monic quadratics using the sum/product pair method.
- Day 5: factorise quadratics whose leading coefficient is greater than one.
- Day 6: solve an unseen mixed set, keeping expansion as a final check.
- Day 7: retest prior mistakes with changed coefficients rather than copied examples.
Frequently asked questions
What is the difference between factorising and solving a quadratic?
Factorising rewrites an expression into a product of factors. Solving finds the value or values of the variable that satisfy an equation. The zero-product rule can solve a factorised quadratic equation when the product is equal to zero.
Must a G2 learner use the quadratic formula for every quadratic?
No. The 2027 K210 syllabus covers factorisation within N5 and, within N7, methods for solving quadratics including factorisation, the formula, completing the square and graphs. Students should know the applicable methods without turning every short factorisation into an unnecessarily long calculation.
How does factorisation help with algebraic fractions?
Factorisation can expose common multiplicative factors in a numerator and denominator. Those factors may be simplified as long as original zero-denominator restrictions are preserved. Read the separate algebraic fractions guide for those checks.
What should a Bukit Timah parent ask a Mathematics tutor?
Ask how the tutor distinguishes sign mistakes, missing cross-products and misunderstanding the sum/product relationship. Then ask how the student is retested independently after a delay. Confirm actual Secondary 4 G2 timetable, tutor and price with the provider; this article does not imply an open seat.
Continue the Secondary 4 G2 Mathematics series
- Changing the subject of formulae
- Equations and simultaneous solutions
- Algebraic fractions
- Exponential and power graphs
- 2026 N(A) and 2027 SEC preparation
Official reading: 2026 N(A)-Level subject directory · 2027 SEC G2 subject directory · 2027 G2 Mathematics K210 syllabus. For the published tutorial approach, read the original unchanged eduKateSG Mathematics small-groups page or enquire through eduKate Singapore. The worked questions are original learning material, not official SEAB examination questions.