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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Dividing by (ax − b): the remainder is P(b/a), and (ax − b) is a factor when P(b/a) = 0.
Two pieces of information (a factor and a remainder) give two equations for two unknown coefficients.
To divide, use long division, synthetic division or compare coefficients: P(x) = (x − a)(px² + qx + r).
For a polynomial with integer coefficients, any integer root divides the constant term, so test those first.
Sum and product of roots: for axⁿ + bxⁿ⁻¹ + … + c, the sum of the roots is −b/a and the product is (−1)ⁿc/a.
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.
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.