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S3 Maths: Factorising Quadratic Expressions (Paper 3 (challenge))
Tutor Prep Guide · Taking out common factors, difference of two squares, perfect squares and cross multiplication, factorising quadratic trinomials over the integers and checking the result
S3 Maths: Factorising Quadratic Expressions (Paper 3 (challenge))
Paper 3 (challenge): ten questions on factorising quadratic expressions, with answers and one line notes.
Direct answers Level: S3 maths; format: factorisation questions using common factors, difference of squares, perfect squares and cross multiplication.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S3 Maths: Factorising Quadratic Expressions |
| Level | S3 |
02 Questions (10 items)
- FoundationFactorise 4x squared minus 12x plus 9.
- FoundationFactorise 25x squared minus 49.
- FoundationFactorise 3x cubed minus 12x completely.
- IntermediateFactorise 6x squared plus 7x minus 3.
- IntermediateFactorise 4x squared minus 4x minus 15.
- IntermediateFactorise x to the fourth minus 16 completely.
- ChallengeFactorise 8x squared minus 14x plus 3.
- ChallengeFactorise 2x squared plus 5xy plus 3y squared.
- ChallengeA rectangle has area x squared plus 7x plus 12 and is 1 unit longer than it is wide. Find the area when x is 5.
- Error redoError clinic: a student expands (2x + 3) squared as 4x squared plus 9. What is the correct expansion?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | (2x – 3) squared | The first term is (2x) squared and the last is 3 squared, and twice 2x times 3 gives the 12x middle term. |
| 2 | (5x + 7)(5x – 7) | 25x squared is (5x) squared and 49 is 7 squared, so the difference of squares applies. |
| 3 | 3x(x + 2)(x – 2) | Take out 3x to get x squared minus 4, then factorise the difference of squares. |
| 4 | (3x – 1)(2x + 3) | Expanding gives 6x squared plus 9x minus 2x minus 3, and the middle terms add to 7x. |
| 5 | (2x – 5)(2x + 3) | Expanding gives 4x squared plus 6x minus 10x minus 15, and the middle terms add to minus 4x. |
| 6 | (x squared + 4)(x + 2)(x – 2) | Treat it first as a difference of squares giving (x squared + 4)(x squared – 4), then factorise x squared minus 4. |
| 7 | (4x – 1)(2x – 3) | Expanding gives 8x squared minus 12x minus 2x plus 3, and the middle terms add to minus 14x. |
| 8 | (2x + 3y)(x + y) | Expanding gives 2x squared plus 2xy plus 3xy plus 3y squared, and the middle terms add to 5xy. |
| 9 | 72 | The area factorises as (x + 3)(x + 4), so the width is x + 3 and the length x + 4, giving 8 times 9 equal to 72. |
| 10 | 4x squared plus 12x plus 9 | The square of a sum needs the middle term twice 2x times 3, which is 12x, and dropping it is the most common error. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Take out any common factor first, then count the terms: two terms point to a difference of squares, three terms to a perfect square or cross multiplication. Always expand the answer to check the middle term.Tutor Registration →
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