S4 Maths: Remainder Theorem and Factor Theorem (Paper 4 (error clinic))
- 2026-11-15
S4 Maths: Remainder Theorem and Factor Theorem (Paper 4 (error clinic))
Paper 4 (error clinic): ten questions on the remainder and factor theorem with answers and one-line notes.
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: S4 Maths: Remainder Theorem and Factor Theorem |
| Level | S4 |
02 Questions (10 items)
- FoundationLet P(x) = x^2 – 4x + 3. Find the remainder when P(x) is divided by (x – 3).
- FoundationLet P(x) = x^3 + 1. Find the remainder when P(x) is divided by (x + 1).
- FoundationLet P(x) = 3x^2 – 5x + 2. Decide whether (x – 1) is a factor of P(x).
- IntermediateLet P(x) = x^3 – 2x^2 – 5x + 6. Find the remainder when P(x) is divided by (x – 2).
- IntermediateLet P(x) = x^3 + kx + 4. Given that (x + 2) is a factor, find k.
- IntermediateLet P(x) = x^3 – 6x^2 + 11x – 6. Find the remainder when P(x) is divided by (x – 1).
- ChallengeLet P(x) = 2x^3 – 3x^2 – 11x + 6. Find the remainder when P(x) is divided by (2x + 1).
- ChallengeLet P(x) = x^3 + 2x^2 – x – 2. Factorise P(x) completely.
- ChallengeLet P(x) = x^3 + ax^2 + bx – 10. Given that (x – 1) and (x + 2) are both factors, find a and b.
- Error redoError clinic: a student finds the remainder of x^2 + x – 6 divided by (x + 3) and writes P(3) = 6. What is the correct answer?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | The remainder is P(3) = 0. | Dividing by (x – 3) means substituting x = 3; a zero remainder makes (x – 3) a factor. |
| 2 | The remainder is P(-1) = 0. | Cubing -1 gives -1, which cancels the constant term +1. |
| 3 | Yes. P(1) = 0. | P(1) = 0 and “(x – 1) is a factor” say exactly the same thing. |
| 4 | The remainder is P(2) = -4. | A non-zero remainder means the division is not exact and (x – 2) is not a factor. |
| 5 | P(-2) = -8 – 2k + 4 = 0, so k = -2. | Watch the cube of a negative number and the final division by a negative. |
| 6 | The remainder is P(1) = 0. | The zero remainder makes (x – 1) a factor; here (x – 2) and (x – 3) are factors too. |
| 7 | Solving 2x + 1 = 0 gives x = -1/2, so the remainder is P(-1/2) = 21/2. | With a fraction, evaluate the square and cube first, add, and simplify last. |
| 8 | P(x) = (x – 1)(x + 1)(x + 2). | Testing P(1) = 0 gives the factor (x – 1); the quotient x^2 + 3x + 2 then splits further. |
| 9 | a = 6 and b = 3. | Write one equation from P(1) = 0 and one from P(-2) = 0; eliminating b gives 3a = 18. |
| 10 | Substitute x = -3: P(-3) = 0. | A positive constant inside the divisor means substituting a negative number; this is the commonest sign slip. |
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.
For remainder and factor questions, first read the divisor carefully. Dividing by (x – a) means substituting x = a, dividing by (x + a) means substituting x = -a, and dividing by (ax – b) means substituting x = b/a: one theorem, three different equations to solve. Only when the remainder is zero may you write that (x – a) is a factor. The other way round, when a question states that an expression is a factor, set P(a) = 0 and solve for the unknown constant, using one equation for each condition and solving them together, then substituting the answer back to check. After finding one factor, use long division or compare terms to obtain the quotient and keep factorising until nothing further splits. Finally check that each root is paired with the right sign: P(2) = 0 goes with (x – 2) and P(-2) = 0 goes with (x + 2).Tutor Registration →
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