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S5 Maths: Remainder and Factor Theorem: Paper 2 (Intermediate)
S5 Maths: Remainder and Factor Theorem: Paper 2 (Intermediate)
Unknown coefficients, quadratic divisors and reading function values from a remainder.
Direct answers Items 1-3: 2, 6, k = 1. Items 4-6: a = -6 b = 11, p = -1 q = -6, 4x + 6. Items 7-9: 2 and -7, x + 1, -1/2. Item 10: k = 2.
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: S5 Maths: Remainder and Factor Theorem |
| Level | Intermediate |
02 Questions (10 items)
- FoundationFind the remainder when x^3 – 2x^2 + 3x – 4 is divided by x – 2.
- FoundationFind the remainder when 2x^3 + 3x^2 – x + 5 is divided by 2x + 1.
- FoundationThe remainder of x^3 + kx^2 – 4x + 6 divided by x + 2 is 10. Find k.
- IntermediateGiven that x^3 + ax^2 + bx – 6 has factors x – 1 and x – 2, find a and b.
- IntermediateGiven f(x) = x^3 + px + q has factor x – 2 and f(1) = -6, find p and q.
- IntermediateFind the remainder when x^3 + 2x^2 + 3x + 4 is divided by x^2 – 1.
- ChallengeThe remainder of f(x) divided by (x – 1)(x + 2) is 3x – 1. Find f(1) and f(-2).
- Challengex^3 – 4x^2 + x + 6 has factors x – 2 and x – 3. Find the third factor.
- ChallengeFind the remainder when 2x^3 – 5x^2 + 4 is divided by 2x – 3.
- Error redoFor x^3 + kx + 2 divided by x – 1 the remainder is 5, but a student gets k = -3. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 2 | Put x = 2: 8 – 8 + 6 – 4 = 2. |
| 2 | 6 | Put x = -1/2: -0.25 + 0.75 + 0.5 + 5 = 6. |
| 3 | k = 1 | Put x = -2: -8 + 4k + 8 + 6 = 4k + 6 = 10, so k = 1. |
| 4 | a = -6, b = 11 | Setting f(1) = 0 and f(2) = 0 gives a = -6 and b = 11. |
| 5 | p = -1, q = -6 | f(2) = 0 gives 2p + q = -8; subtracting p + q = -7 leaves p = -1, so q = -6. |
| 6 | 4x + 6 | Replace x^2 by 1: x + 2 + 3x + 4 = 4x + 6. |
| 7 | f(1) = 2, f(-2) = -7 | The remainder agrees at the roots: f(1) = 3 – 1 = 2 and f(-2) = -6 – 1 = -7. |
| 8 | x + 1 | With leading coefficient 1, 6 divided by (-2)(-3) is 1, so the third factor is x + 1. |
| 9 | -1/2 | Put x = 3/2: 27/4 – 45/4 + 16/4 = -2/4 = -1/2. |
| 10 | The remainder is not zero; k = 2 | 1 + k + 2 = 5 gives k = 2; treating the remainder as zero produces the wrong -3. |
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.
Set the divisor to zero, substitute that x into the polynomial and compare with zero.Tutor Registration →
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