S4 Maths: Remainder Theorem and Factor Theorem (Paper 2 (intermediate))

S4 Maths: Remainder Theorem and Factor Theorem (Paper 2 (intermediate))

Paper 2 (intermediate): ten questions on the remainder and factor theorem with answers and one-line notes.

01　Paper details (check answers afterwards)

02　Questions (10 items)

1. FoundationLet P(x) = x^3 – 7x + 6. Find the remainder when P(x) is divided by (x – 2).
2. FoundationLet P(x) = 3x^3 + 2x^2 – 7x + 2. Find the remainder when P(x) is divided by (x – 1).
3. FoundationLet P(x) = x^3 + 4x^2 + x – 6. Find the remainder when P(x) is divided by (x + 2).
4. IntermediateLet P(x) = 2x^3 – 5x^2 + ax – 3 with P(1) = 0. Find a.
5. IntermediateLet P(x) = x^3 + ax^2 – 4x + b. Given that (x – 1) is a factor and the remainder on dividing by (x + 2) is 6, find a and b.
6. IntermediateLet P(x) = 2x^3 + x^2 – 13x + 6. Given that (x – 2) is a factor, factorise P(x) completely.
7. ChallengeLet P(x) = x^4 – 2x^3 – 3x^2 + 4x + 4. Find the remainder on dividing by (x + 1) and decide whether (x + 1) is a factor.
8. ChallengeLet P(x) = x^3 – 2x^2 + 5x – 7. Find the remainder when P(x) is divided by (2x + 1).
9. ChallengeLet P(x) = x^3 + ax^2 + bx – 6. Given that both (x – 1) and (x + 2) are factors, find a and b.
10. Error redoError clinic: after finding P(2) = 0 a student writes that (x + 2) is a factor. What is the correct statement?

03　Answers and one-line notes

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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