HL AA · Functions

Polynomial division and the factor theorem (IB Math AA HL)

Updated October 2026

Quick answer: The remainder theorem: the remainder when P(x) is divided by (x − a) is P(a). The factor theorem: (x − a) is a factor exactly when P(a) = 0. To factorise a cubic, find one root by trying small values (or use a given one), divide by that factor, then factorise the quadratic that is left.

P(a) = 0 iff (x-a) is a factor. HL AA loves 'given that x = 2 is a root of the cubic, factorise fully' — do it by inspection or synthetic division.

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Key ideas

Worked example

P(x) = 2x³ + ax² + bx − 6. (x − 2) is a factor of P(x), and the remainder when P(x) is divided by (x + 1) is 6. Find a and b, and factorise P(x) fully.

  1. P(2) = 16 + 4a + 2b − 6 = 0, so 2a + b = −5.
  2. P(−1) = −2 + a − b − 6 = 6, so a − b = 14.
  3. Add the equations: 3a = 9, so a = 3 and b = −11.
  4. Divide by (x − 2): 2x³ + 3x² − 11x − 6 = (x − 2)(2x² + 7x + 3).
  5. 2x² + 7x + 3 = (2x + 1)(x + 3), so P(x) = (x − 2)(2x + 1)(x + 3).

Answer: a = 3, b = −11; P(x) = (x − 2)(2x + 1)(x + 3).

Exam-style HL AA questions on this topic

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5 quick questions on the factor and remainder theorems. No sign-in needed.

Find the remainder when \(x^{3} + 2 x^{2} + 2 x + 1\) is divided by \((x + 2)\).
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FAQ

What is the difference between the factor theorem and the remainder theorem?

The remainder theorem says the remainder on dividing P(x) by (x − a) is P(a). The factor theorem is the special case where that remainder is 0: then (x − a) is a factor of P(x).

Where can I practise polynomial division and the factor theorem questions?

On the Polynomials and Rational Functions questions page: 79 exam-style HL AA questions with full mark schemes, plus free step-by-step worked examples.

Can I use my calculator for polynomial division and the factor theorem?

In HL AA, Paper 1 is non-calculator and the other papers allow a GDC, so learn the exact method as well as the calculator one.

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