Polynomials & Partial Fractions (AA HL)
This topic covers two closely linked skills: using the factor and remainder theorems (plus Vieta's formulas) to understand a polynomial's roots directly from its coefficients, and splitting a fraction with a quadratic denominator back into two simpler fractions. Both techniques turn up constantly elsewhere in the AA HL course, especially in integration.
What the syllabus says
This topic maps onto two points in the official IB Analysis & Approaches syllabus.
| Code | Syllabus content |
|---|---|
| AHL1.11 | Partial fractions, with a maximum of two distinct linear terms in the denominator, and the degree of the numerator less than the degree of the denominator. |
| AHL2.12 | Polynomial functions, their graphs and equations; zeros, roots and factors. The factor and remainder theorems. Sum and product of the roots of a polynomial equation. |
Partial fractions links forward to integrating rational functions (AHL5.15); sum and product of roots links to powers and roots of complex numbers (AHL1.14).
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is the remainder theorem?
The remainder theorem says that when a polynomial \(p(x)\) is divided by \((x-a)\), the remainder is exactly \(p(a)\) - no division is actually needed to find it, just substitution. It's a shortcut that turns a division problem into an evaluation problem.
e.g. For \(p(x)=3x^3-2x^2+x-7\) divided by \((x+1)\): remainder \(=p(-1)=-3-2-1-7=-13.\)
What is the factor theorem?
The factor theorem is the special case of the remainder theorem where the remainder is zero: if \(p(a)=0\), then \((x-a)\) is a factor of \(p(x)\). It's how you find factors of a cubic or quartic without guessing-and-checking by long division.
e.g. For \(p(x)=x^3+2x^2-5x-6\): \(p(-3)=-27+18+15-6=0\), so \((x+3)\) is a factor.
What do "sum and product of roots" mean?
For a polynomial equation \(a_nx^n+\cdots+a_1x+a_0=0\) with roots \(\alpha,\beta,\gamma,\ldots\), the sum and product of all the roots can be read straight off the coefficients, without solving the equation at all - useful whenever a question only asks about a combination of the roots.
e.g. For \(x^3-6x^2+11x-6=0\): sum of roots \(=-(-6)/1=6\); product of roots \(=-(-6)/1=6.\)
What are partial fractions?
Partial fractions is the reverse of adding fractions together - it splits one fraction with a factorised denominator back into a sum of simpler fractions, each with just one linear factor on the bottom. This makes the expression much easier to integrate or expand as a series.
e.g. \(\dfrac{7x-1}{(x-1)(x+2)} = \dfrac{2}{x-1}+\dfrac{5}{x+2}.\)
What is a repeated substitution (cover-up) value?
When solving for the unknown constants in a partial fraction identity, substituting \(x\) equal to a root of one of the denominator's factors makes one term vanish completely, instantly giving the other constant. It avoids comparing coefficients term by term.
e.g. In \(4x+7=A(x+3)+B(x+1)\), setting \(x=-1\) gives \(3=2A\), so \(A=\tfrac32\) instantly.
Key formulas
Five results cover almost every question on this topic. The two tables below summarise all of them at a glance - the explanations underneath go into more depth on each one.
Formula reference
Sum and product of roots are on the official formula booklet; the factor/remainder theorems and the partial-fraction setup are treated as prior-knowledge methods.
| Formula | Used for | Booklet? |
|---|---|---|
| \(p(x)=(x-a)q(x)+p(a)\) | Remainder theorem | Not in booklet - prior knowledge |
| \(p(a)=0 \iff (x-a)\) is a factor | Factor theorem | Not in booklet - prior knowledge |
| For \(\displaystyle\sum_{r=0}^{n}a_rx^r=0\): sum of roots \(=-\dfrac{a_{n-1}}{a_n}\) | Sum of the roots | ✓ Yes |
| Product of roots \(=\dfrac{(-1)^na_0}{a_n}\) | Product of the roots | ✓ Yes |
| \(\dfrac{px+q}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}\) | Partial fractions, distinct linear factors | Not in booklet - method |
Factor theorem vs remainder theorem
They're two sides of the same idea - one is a special case of the other.
| Feature | Remainder theorem | Factor theorem |
|---|---|---|
| What it tells you | The remainder after dividing \(p(x)\) by \((x-a)\) | Whether \((x-a)\) is a factor of \(p(x)\) |
| Condition | Remainder \(=p(a)\), for any \(a\) | \(p(a)=0\) specifically |
| Typical use | Find a remainder without doing the division | Find roots to factorise a cubic or quartic fully |
The factor and remainder theorems
Both theorems come from the same division statement: \(p(x)=(x-a)q(x)+p(a)\).
Remainder theorem
Substitute \(x=a\) directly into \(p(x)\) to get the remainder on division by \((x-a)\) - no long division needed.
Factor theorem
If \(p(a)=0\), then \((x-a)\) is a factor. Test small integer factors of the constant term first.
Full factorisation
Once one factor is found, divide to get a lower-degree quotient, then repeat or factorise the quotient directly.
Sum and product of roots
Vieta's formulas let you answer questions about combinations of roots without ever solving the equation.
Vieta for a cubic
\[\alpha+\beta+\gamma=-\frac{a_2}{a_3},\ \ \alpha\beta\gamma=-\frac{a_0}{a_3}\]
Read the sum and product of the three roots straight off the coefficients.
Symmetric functions
\[\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)\]
Combinations like sums of squares can be rebuilt from the standard sums using algebraic identities.
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
Find the remainder when \(p(x)=3x^3-2x^2+x-7\) is divided by \((x+1)\).
Worked solution
By the remainder theorem the remainder is \(p(-1)\). M1
\(p(-1)=-3-2-1-7=-13\). A1
Express \(\dfrac{7x-1}{(x-1)(x+2)}\) in partial fractions.
Worked solution
\(\dfrac{7x-1}{(x-1)(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{x+2}.\) M1
\(7x-1=A(x+2)+B(x-1).\) \(x=1:\ 6=3A\Rightarrow A=2;\) A1
\(x=-2:\ -15=-3B\Rightarrow B=5.\) A1
So \(\dfrac{2}{x-1}+\dfrac{5}{x+2}.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- 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 \(x=a\).
- 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.
- 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.
- Not checking the partial fraction identity holds for all \(x\). After finding \(A\) and \(B\), it's worth checking with one more value of \(x\) (or by comparing a coefficient) that the identity actually balances.
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
The most common arithmetic errors on the GDC come from missing brackets - especially with fractions, negatives and powers, which matters when entering a polynomial like \(3x^3-2x^2+x-7\) or checking a value of \(p(a)\).
- Use brackets whenever you have a fraction, a negative, or a compound expression in an exponent.
- Fraction: type (3+5)÷(2−1) not 3+5÷2−1 - the calculator respects order of operations, so division binds tightly.
- Power: type (2x+1)^3 not 2x+1^3 - without brackets only the 1 is cubed.
- Square root: the √ key only captures what immediately follows - use √(expression) with a closing bracket for multi-term expressions.
- Powers: use the ^ key. Square root: 2nd → √ then close the bracket. Cube root: MATH → 4:∛( or x^(1/3).TI-84
- Powers: use the ^ key or the exponent template. Roots: use the √ template from the maths palette (ctrl+B or the template key).Nspire
- Powers: use the ^ key (x□ key). Square root: SHIFT → √. Fractions: use the fraction template (SHIFT → ÷) for clean stacked fractions.Casio
- Always check the answer makes sense - a wildly large or small result usually means a missing bracket.
Tip: When in doubt, add extra brackets. (3+5)÷(2) and 3+5÷2 give different answers - the first is almost always what you mean.
For very large or very small numbers - avoids typing long strings of zeros and prevents rounding errors, useful when a polynomial evaluation or a partial-fractions integral produces a large result.
- Scientific notation means \(a \times 10^n\), e.g. \(3.2 \times 10^8\) or \(4.5 \times 10^{-3}.\)
- Use 2nd → , (EE) to enter the ×10 part: type 3.2 2nd , 8 to enter \(3.2\times10^8.\) Do NOT type ×10^ separately.TI-84
- Use the EE key (or type ×10^ from the keyboard template). Or just type 3.2×10^8 using the ^ key.Nspire
- Use the ×10ˣ key (EXP key) - type 3.2 then EXP then 8. Do NOT type ×10^ manually.Casio
- To display answers in scientific notation: on TI-84 press MODE and choose SCI; on Casio set the display mode in SET UP.
Tip: A common mistake is typing ×10^ instead of using the EE/EXP key - this gives ×10×... (multiplication, then a power) rather than proper scientific notation.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Polynomials & partial fractions questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between the factor theorem and the remainder theorem?
The remainder theorem tells you the remainder when \(p(x)\) is divided by \((x-a)\) is \(p(a)\), for any \(a\). The factor theorem is the special case where that remainder is zero - if \(p(a)=0\), then \((x-a)\) is a factor of \(p(x)\).
Why does partial fractions only work with distinct linear factors here?
The AA syllabus caps this topic at a maximum of two distinct linear terms in the denominator, with the numerator's degree less than the denominator's. Repeated factors and irreducible quadratic factors in the denominator are not required.
How do I remember Vieta's formulas for sum and product of roots?
For \(a_nx^n+\cdots+a_1x+a_0=0\), the sum of the roots is \(-a_{n-1}/a_n\) (the second coefficient, with a sign flip, over the leading one), and the product of the roots is \((-1)^na_0/a_n\) (the constant term over the leading one, with a sign that alternates with the degree).
Can I use my GDC for this topic?
Yes, on Paper 2 - you can graph a polynomial and read off its roots directly, or use a polynomial root finder. On Paper 1 you need the factor theorem, remainder theorem and partial fractions by hand. See the GDC guide for model-specific instructions.
Sub-topics
Polynomials & Partial Fractions broken down into its individual skills, each with its own focused page.
Related topics
More Number & Algebra topics from the same AA HL syllabus unit, in case you want to keep going.