Polynomial Division & Factor Theorem (AA HL)
Once a polynomial gets past degree 2, factorising it by inspection stops being realistic - the factor theorem gives you a systematic way in, and polynomial division (or the sum/product of roots shortcut) finishes the job. This page covers testing and finding factors, dividing out a known factor, and reading off root sums and products directly from the coefficients, with worked examples and the mistakes that lose the most marks. It's part of the broader Polynomials & Partial Fractions topic.
54 questions on this sub-topic.
The factor theorem and root sums
Covered under IB syllabus reference AHL2.12: polynomial functions, their graphs and equations, zeros, roots and factors, the factor and remainder theorems, and the sum and product of the roots of a polynomial equation found directly from its coefficients.
Factor theorem
\(p(a)=0 \iff (x-a)\) is a factor
Not in the formula booklet - prior knowledge. Test small integer factors of the constant term first.
Remainder theorem
\(p(x)=(x-a)q(x)+p(a)\)
Not in the formula booklet - prior knowledge. The remainder on dividing by \((x-a)\) is just \(p(a)\), no division required.
Need the full syllabus wording and formula-booklet reference table? See Polynomials & Partial Fractions. For GDC-based root finding, see the parent topic's GDC guidance.
Worked examples
Find the remainder when \(p(x) = 2x^3 - 3x^2 + x - 5\) is divided by \((x - 2).\)
Worked solution
By the remainder theorem the remainder is \(p(2).\) M1
\(p(2) = 2(8) - 3(4) + 2 - 5 = 16 - 12 + 2 - 5\)
\(= 1.\) A1
The roots of \(2x^2 - 5x + 1 = 0\) are \(\alpha, \beta\).
Find \(\alpha + \beta\), \(\alpha\beta\), then \(\alpha^2 + \beta^2\).
Worked solution
For \(ax^2+bx+c=0\): \(\alpha+\beta = -\tfrac{b}{a}\) and \(\alpha\beta = \tfrac{c}{a}\). M1
With \(a=2,\ b=-5,\ c=1\): \(\alpha+\beta = \tfrac{5}{2},\ \alpha\beta = \tfrac{1}{2}.\) A1
\(\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta\): M1
\(\alpha^2+\beta^2 = \left(\tfrac{5}{2}\right)^2 - 2\cdot\tfrac{1}{2}\) A1
\(= \tfrac{25}{4} - 1 = \tfrac{21}{4}.\) A1
Common mistakes
- Stopping after finding one factor. The factor theorem only finds a single root at a time - after dividing out that factor, keep factorising (or applying the theorem again) until the polynomial is fully broken down.
- Testing \(p(-a)\) instead of \(p(a)\) for the factor \((x+a)\). \((x+a)\) is really \((x-(-a))\), so the theorem requires substituting \(x=-a\), not \(x=a\).
- Forgetting the sign in the product of roots. The product of roots is \((-1)^na_0/a_n\) - the alternating sign is easy to drop, especially for even-degree polynomials.
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55 polynomial division and factor theorem questions, marked instantly like the real exam.
Quick answers
What is the factor theorem?
If \(p(a) = 0\) for a polynomial \(p(x)\), then \((x-a)\) is a factor of \(p(x)\). It lets you test candidate factors quickly, usually by trying small integer factors of the constant term first.
How do you find the sum and product of the roots of a polynomial?
Straight from the coefficients: for a degree-\(n\) polynomial with leading coefficient \(a_n\) and constant term \(a_0\), the sum of the roots is \(-\tfrac{a_{n-1}}{a_n}\) and the product of the roots is \((-1)^n\tfrac{a_0}{a_n}\).