CSEC ADDITIONAL MATHEMATICS · SECTION 1 · LESSON A2

The Remainder Theorem and the Factor Theorem

Finding a remainder without dividing, testing for factors, factorising cubics, and using the theorems to find unknown coefficients.

Algebra, Sequences and SeriesS1 A.3S1 A.4S1 A.5≈ 40 min🎮 3 games
Before you start

By the end of this lesson you can…

Step 1 · Watch

Explainer video

5 narrated scenes. Press play, listen and read along — the points appear as they are explained. Use ‹ Prev / Next › or the scene bar to jump to any scene.

LESSON A2 · EXPLAINER

The Remainder Theorem and the Factor Theorem

“What is the remainder when this cubic is divided by x − 3?” You could divide — or you could just evaluate f(3). One number, one line.

Narration uses your device’s built-in voice. No sound? Switch to “Captions only” — the captions advance on their own.

Explore in 3D

Remainder and factor theorem

Test a value of x. If f(k) = 0, then (x − k) is a factor.

Step 2 · Read

Key notes

Copy these into your notebook in your own words — that is how they stick.

Remainder Theorem

Remainder on division by (x − a) is f(a). By (ax − b): f(b/a).

Factor Theorem

(x − a) is a factor ⇔ f(a) = 0.

Finding roots

Try ±factors of the constant term (and ±factor ratios when the leading coefficient is not 1).

Factorising

Root → divide → factorise the quotient.

Unknowns

Each condition gives one equation: factor → f(a) = 0; remainder r → f(a) = r.

Check

Substitute the final coefficients back into f(a).

Link

The remainder when f(x) is divided by a quadratic is linear: Ax + B.

Notation

f(a) means replace every x by a and evaluate.

EXAM TIPS & COMMON MISTAKES

  • Show f(a) = … = 0 explicitly when asked to “show that (x − a) is a factor”.
  • For (x + 2) use a = −2: the sign flips.
  • Do the arithmetic carefully with negative powers: (−2)³ = −8.
  • After finding unknowns, state the final polynomial and check it with one more substitution.
Play & practise

Games

3 quick games on this lesson. Earn XP and badges as you play — your best score for each game is saved. Games are for practice and do not change your exam readiness. More games in the Additional Mathematics Arcade.

Step 3 · Practise

Paper 1 practice — multiple choice

8 items modelled on real CSEC Paper 1 questions. Choose one answer for each, then check. Aim for 80% or better. Work each item on paper first.

CSEC Paper 01 style · SO A.3

1. The remainder when f(x) = 2x³ − 5x² + x − 7 is divided by x − 3 is

Answer: D. f(3) = 54 − 45 + 3 − 7 = 5.
CSEC Paper 01 style · SO A.3

2. The remainder when x³ + 4x − 5 is divided by x + 1 is

Answer: A. f(−1) = −1 − 4 − 5 = −10.
CSEC Paper 01 style · SO A.4

3. Which of these is a factor of x³ + 3x² − 4x − 12?

Answer: B. f(−3) = −27 + 27 + 12 − 12 = 0.
CSEC Paper 01 style · SO A.4

4. x³ + 3x² − 4x − 12 factorised completely is

Answer: C. Roots 2, −2, −3.
CSEC Paper 01 style · SO A.5

5. (x − 1) is a factor of x³ + kx² − 7x + 6. The value of k is

Answer: D. f(1) = 1 + k − 7 + 6 = k = 0.
CSEC Paper 01 style · SO A.5

6. x³ + ax² + bx + 6 has factor (x − 1) and remainder 12 on division by (x − 2). The value of a is

Answer: A. a + b = −7 and 2a + b = −1 give a = 6.
CSEC Paper 01 style · SO A.4

7. For (2x − 1) to be a factor of f(x), which value must be tested?

Answer: B. Put 2x − 1 = 0, so x = 1/2.
CSEC Paper 01 style · SO A.5

8. If f(2) = 0 for f(x) = 2x³ − 5x² + kx + 6, then k is

Answer: C. 16 − 20 + 2k + 6 = 0 → k = −1.
Step 4 · Apply

Paper 2 practice — structured questions

Write full answers, then reveal the mark scheme and tick the points you earned (24 marks in total). Show all your working — method marks count. Your answers are saved on this device.

CSEC Paper 02 style · Section 1 (Factor and Remainder Theorems)

Question 1 [13 marks]

f(x) = x³ + ax² + bx + 6. When f(x) is divided by x − 2 the remainder is 12 and x − 1 is a factor.

(a) Show that a + b = −7 and 2a + b = −1. [4]

MARK SCHEME — (a) (max 4)

(b) Hence find a and b. [2]

MARK SCHEME — (b) (max 2)

(c) Factorise f(x) completely. [4]

MARK SCHEME — (c) (max 4)

(d) Solve f(x) = 0 correct to 2 decimal places. [3]

MARK SCHEME — (d) (max 3)

CSEC Paper 02 style · Section 1 (Cubic with a factor)

Question 2 [11 marks]

p(x) = 2x³ − 5x² + kx + 6 has x − 2 as a factor.

(a) Show that k = −1. [2]

MARK SCHEME — (a) (max 2)

(b) Factorise p(x) completely. [4]

MARK SCHEME — (b) (max 4)

(c) Find the remainder when p(x) is divided by x + 2. [2]

MARK SCHEME — (c) (max 2)

(d) Hence, or otherwise, solve p(x) > 0. [3]

MARK SCHEME — (d) (max 3)

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Sources

Where this topic appears in CSEC papers

Find these in your own past-paper booklets for extra timed practice. Items above are written in the style of these papers; any adapted from a specific question say so.