中文（繁體）
中文（简体）
English
S5 Maths: Remainder and Factor Theorem: Paper 3 (Challenge)
S5 Maths: Remainder and Factor Theorem: Paper 3 (Challenge)
Repeated factors, divisors of degree three or more, quadratic fitting and simultaneous conditions.
Direct answers Items 1-3: a = 0, b = 2, c = 3; yes; x + 8. Items 4-6: p = 0, q = -7; 3x – 1; a = 2, b = -8. Items 7-9: 3; a = 4, b = -3; x – 1. Item 10: x + 1 is not a factor.
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 | Challenge |
02 Questions (10 items)
- Foundationf(x) = x^3 + ax^2 + bx + c has f(1) = 6, f(-1) = 0 and f(2) = 15. Find a, b, c.
- FoundationIs (x – 1)^2 a factor of x^3 – 3x + 2?
- FoundationFind the remainder when 2x^4 + 3x^3 – x + 5 is divided by x^3 – 1.
- Intermediatex^3 + px^2 + qx + 6 is divisible by both x – 1 and x + 3. Find p and q.
- Intermediatef(x) leaves remainder 5 on division by x – 2 and -4 by x + 1. Find the remainder for (x – 2)(x + 1).
- Intermediate2x^3 + ax^2 – 8x + b is divisible by x – 2 and x + 1. Find a and b.
- ChallengeA quadratic f(x) has f(1) = 3, f(2) = 5 and f(3) = 9. Find f(0).
- Challengex^3 – 2x^2 + ax + b has factor x – 1 and f(2) = 5. Find a and b.
- ChallengeFind the remainder when x^4 – 1 is divided by x^2 + x + 1.
- Error redoA student sees f(1) = 0 and claims x + 1 is a factor of f(x) = x – 1. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | a = 0, b = 2, c = 3 | Solving the three equations gives a = 0, b = 2, c = 3, so f(x) = x^3 + 2x + 3. |
| 2 | Yes: f(1) = 0 and the derivative also vanishes at x = 1 | A repeated factor needs two conditions: f(1) = 1 – 3 + 2 = 0 and 3x^2 – 3 is also zero at x = 1. |
| 3 | x + 8 | Replace x^3 by 1: 2x + 3 – x + 5 = x + 8. |
| 4 | p = 0, q = -7 | f(1) = 0 gives p + q = -7 and f(-3) = 0 gives 3p – q = 7, so p = 0 and q = -7. |
| 5 | 3x – 1 | Let the remainder be ax + b: 2a + b = 5 and -a + b = -4 give a = 3, b = -1. |
| 6 | a = 2, b = -8 | f(2) = 0 gives 4a + b = 0 and f(-1) = 0 gives a + b = -6, so a = 2 and b = -8. |
| 7 | 3 | Fitting ax^2 + bx + c gives a = 1, b = -1, c = 3, so f(0) = 3. |
| 8 | a = 4, b = -3 | f(1) = 0 gives a + b = 1 and f(2) = 5 gives 2a + b = 5, so a = 4 and b = -3. |
| 9 | x – 1 | Using x^3 = 1, the remainder of x^4 – 1 reduces to x – 1. |
| 10 | Check f(-1) = -2, so x + 1 is not a factor; f(1) = 0 matches x – 1 | The factor theorem needs the value that makes the divisor zero, so x + 1 requires x = -1. |
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 →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.