Polynomial Functions (AA HL)
A cubic or quartic behaves in ways a quadratic never does - it can have three or four roots, some of them complex, and the way those roots relate to the coefficients is exam gold if you know the shortcut. This page covers the factor and remainder theorems and the sum/product-of-roots results, with the worked examples and slip-ups that come up most under HL exam conditions. It's part of the broader Quadratics & Polynomials topic.
10 questions on this sub-topic.
Key results
Covered under IB syllabus reference AHL2.12: polynomial functions, their graphs and equations; zeros, roots and factors; the factor and remainder theorems; the sum and product of the roots of a polynomial equation, read directly from its coefficients.
Factor & remainder theorems
\(P(a)=0 \iff (x-a)\) is a factor of \(P(x)\)
The remainder when \(P(x)\) is divided by \((x-a)\) is \(P(a)\). If that remainder is zero, \((x-a)\) divides \(P(x)\) exactly - this is the factor theorem, and it's how you hunt for roots of a cubic or quartic without guesswork.
Sum & product of roots
\(\alpha+\beta+\gamma=-\dfrac{b}{a},\ \ \alpha\beta\gamma=-\dfrac{d}{a}\)
For a cubic \(ax^3+bx^2+cx+d\), these come straight from the coefficients - no need to solve for the individual roots first. Both are in the formula booklet, generalised to any degree.
Need the full syllabus wording and formula-booklet reference table? See Quadratics & Polynomials.
Worked examples
\(p(x)=2x^3 + ax^2 + bx - 6\) is divisible by \((2x-1)\) and leaves remainder \(-20\) when divided by \((x+1)\).
Find \(a\) and \(b\).
(a)(i) Find \(a\).
(a)(ii) Find \(b\).
Worked solution
\(p(\frac12) = 0\): \(\frac14 + \frac{a}{4} + \frac{b}{2} - 6 = 0 \Rightarrow a + 2b = 23.\) M1 A1
\(p(-1) = -2 + a - b - 6 = -20 \Rightarrow a - b = -12.\) M1 A1
subtract: \(3b = 35 \Rightarrow b = \frac{35}{3},\ a = -\frac13.\) M1 A1 A1
The cubic \(x^3 + ax + b\) (real \(a,b\)) has a root \(x = 1+2i\).
Find \(a\) and \(b\) and the third root.
Worked solution
real coefficients \(\Rightarrow 1 - 2i\) is also a root, giving factor \(x^2 - 2x + 5.\) R1 A1
the \(x^2\) coefficient is 0, so the sum of roots is 0, giving third root \(-2.\) M1 A1
\((x^2 - 2x + 5)(x + 2) = x^3 + x + 10.\) M1 So \(a = 1,\ b = 10.\) A1
Common mistakes
- Assuming complex roots always come in conjugate pairs. That's only guaranteed when every coefficient of the polynomial is real - it doesn't apply to a polynomial with complex coefficients.
- Mixing up the factor and remainder theorems. Evaluating \(P(a)\) always gives you the remainder on division by \((x-a)\); it only tells you \((x-a)\) is a factor when that value happens to be zero.
- Reading off the wrong sign in the sum/product-of-roots formulas. The sum of the roots is \(-b/a\), not \(b/a\) - the leading minus sign is easy to drop when working quickly under exam pressure.
Ready to practise properly?
11 polynomial-function questions, marked instantly like the real exam. GDC guidance for graphing roots is on the full topic page.
Quick answers
What is the difference between the factor theorem and the remainder theorem?
The remainder theorem tells you the remainder when a polynomial \(P(x)\) is divided by \((x-a)\): it equals \(P(a)\). The factor theorem is the special case where that remainder is zero, meaning \((x-a)\) is a factor of \(P(x)\).
How do you find the sum and product of the roots of a polynomial from its coefficients?
For a cubic \(ax^3+bx^2+cx+d\), the sum of the roots is \(-b/a\) and the product is \(-d/a\), read straight from the coefficients without solving for individual roots - this only works cleanly when every coefficient is real.