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S4 Maths: Remainder Theorem and Factor Theorem (Paper 1 (basics))

S4 Maths: Remainder Theorem and Factor Theorem (Paper 1 (basics))

  • 2026-11-15

Tutor Prep Guide · Using substitution to find a remainder, the factor theorem, finding an unknown constant from a known factor, long division for the quotient, complete factorisation and the linear form of a remainder

S4 Maths: Remainder Theorem and Factor Theorem (Paper 1 (basics))

Paper 1 (basics): ten questions on the remainder and factor theorem with answers and one-line notes.

Direct answers　Level: S4 maths; format: remainders, factor tests, unknown constants, long division and factorisation.

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)

  1. FoundationLet P(x) = x^2 + 3x + 2. Find the remainder when P(x) is divided by (x – 2).
  2. FoundationLet P(x) = 2x^3 – x + 5. Find the remainder when P(x) is divided by (x – 1).
  3. FoundationLet P(x) = x^2 – 5x + 6. Decide whether (x – 2) is a factor of P(x).
  4. IntermediateLet P(x) = x^3 – 4x^2 + x + 6. Find the remainder on dividing by (x – 3) and decide whether (x – 3) is a factor.
  5. IntermediateLet P(x) = x^3 + kx – 6. Given that (x – 2) is a factor of P(x), find k.
  6. IntermediateLet P(x) = x^3 + 2x^2 – 5x – 6. Find P(-1) and decide whether (x + 1) is a factor.
  7. ChallengeLet P(x) = x^4 – 3x^2 + 2. Find the remainder when P(x) is divided by (x + 2).
  8. ChallengeLet P(x) = 2x^3 – 3x^2 + 4x – 5. Find the remainder when P(x) is divided by (2x – 1).
  9. ChallengeLet P(x) = 2x^3 + x^2 + kx + 4. Given that (x + 1) is a factor, find k.
  10. Error redoError clinic: a student finds the remainder of x^2 + 2x – 3 divided by (x + 3) and writes P(3) = 12. What went wrong?

03　Answers and one-line notes

# Answer One-line note
1 The remainder is P(2) = 12. The remainder theorem says the remainder on dividing by (x – a) is P(a), so substitute x = 2.
2 The remainder is P(1) = 6. The x^2 term is missing, so treat it as zero when substituting.
3 Yes. P(2) = 0, so the remainder is zero. By the factor theorem, P(a) = 0 exactly when (x – a) is a factor.
4 The remainder is P(3) = 0, so (x – 3) is a factor. The three group values cancel exactly, giving a zero remainder, so the division is exact.
5 P(2) = 8 + 2k – 6 = 0, so k = -1. The factor theorem turns the statement into the equation P(2) = 0, which is then solved for k.
6 P(-1) = 0, so (x + 1) is a factor. The divisor is (x + 1), so substitute x = -1; the sign is the usual trap.
7 The remainder is P(-2) = 6. Even powers stay positive and odd powers turn negative after substituting a negative value.
8 Solving 2x – 1 = 0 gives x = 1/2, so the remainder is P(1/2) = -7/2. With a divisor (ax – b) substitute x = b/a, not a or b.
9 P(-1) = -2 + 1 – k + 4 = 0, so k = 3. Substitute x = -1 first, collect the constants, and only then solve for k.
10 The divisor is (x + 3), so substitute x = -3; the correct remainder is P(-3) = 0. Rewrite the divisor as x – a first; (x + 3) means a = -3, not 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.

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 →

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.

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