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S4 Maths: Remainder Theorem and Factor Theorem (Paper 3 (challenge))
S4 Maths: Remainder Theorem and Factor Theorem (Paper 3 (challenge))
Paper 3 (challenge): 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^3 + kx^2 – 4x + 4. Given that the remainder on dividing by (x – 2) is 8, find k.
- FoundationLet P(x) = 2x^3 + ax^2 + bx – 6. Given that (x – 1) and (x – 2) are both factors, find a and b.
- FoundationLet P(x) = x^3 – 3x^2 – 10x + 24. Given that x = 2 is a root of P(x) = 0, factorise P(x) completely.
- IntermediateLet P(x) = x^3 + ax^2 + bx + 8. Given that (x – 1) and (x + 2) are both factors, factorise P(x) completely.
- IntermediateLet P(x) = 2x^3 + px^2 – 8x + 3. Given that (2x – 1) is a factor, find p and factorise P(x) completely.
- IntermediateLet P(x) = x^4 + ax^3 – 2x + 5. Given that the remainder on dividing by (x + 1) is 10, find a.
- ChallengeLet P(x) = x^3 + ax^2 + bx – 4. The remainder on dividing by (x – 2) is 2 and on dividing by (x + 1) is -4. Find a and b.
- ChallengeLet P(x) = 2x^3 – 5x^2 – 4x + 12. Given that x = 2 is a repeated root, factorise P(x) completely.
- ChallengeA polynomial P(x) leaves remainder 5 on division by (x – 1) and remainder 11 on division by (x – 3). Find the remainder on division by (x – 1)(x – 3).
- Error redoError clinic: to test whether (2x – 1) is a factor of 2x^3 – x^2 – 8x + 4 a student substitutes x = 1, gets P(1) = -3 and concludes it is not a factor. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | P(2) = 8 + 4k – 8 + 4 = 8, so k = 1. | Write out P(2) fully, set it equal to the given remainder, then simplify. |
| 2 | a = -9 and b = 13. | P(1) = 0 gives a + b = 4 and P(2) = 0 gives 2a + b = -5; subtract to obtain a. |
| 3 | P(x) = (x – 2)(x – 4)(x + 3). | A root at 2 gives the factor (x – 2); long division gives x^2 – x – 12, which then splits. |
| 4 | a = -3, b = -6 and P(x) = (x – 1)(x – 4)(x + 2). | Solve P(1) = 0 with P(-2) = 0 for a and b, then divide to reach the third factor. |
| 5 | p = 3 and P(x) = (2x – 1)(x + 3)(x – 1). | A factor of (2x – 1) means P(1/2) = 0; the quotient is then x^2 + 2x – 3. |
| 6 | P(-1) = 1 – a + 2 + 5 = 10, so a = -2. | Missing terms count as zero; substitute every term carefully. |
| 7 | a = 0 and b = -1. | The two equations 2a + b = -1 and a – b = 1 add to give 3a = 0. |
| 8 | P(x) = (x – 2)^2(2x + 3). | A repeated root means the same factor appears twice: after dividing by (x – 2) the quotient still vanishes at 2. |
| 9 | The remainder is 3x + 2. | Dividing by a quadratic leaves a remainder of degree at most one, so write ax + b and use both conditions. |
| 10 | Substitute x = 1/2 instead: P(1/2) = 0, so (2x – 1) is a factor. | A divisor (ax – b) requires x = b/a; substituting 1 only tests the factor (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.
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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