What does the binomial theorem and partial fractions topic cover in O-Level A-Maths 4049?

It covers two related Algebra-strand outcomes in syllabus 4049. The binomial theorem expands $(a + b)^n$ for a positive integer $n$ using binomial coefficients, and the general term lets you pick out a single term such as the coefficient of $x^3$ or the term independent of $x$ without writing the whole expansion. Partial fractions then reverse the process of adding fractions, splitting a proper rational fraction into a sum of simpler ones, with set numerator forms for distinct linear, repeated linear and irreducible quadratic factors.

What is the general term of a binomial expansion and why is it useful?

The general term of $(a + b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^{r}$ for $r = 0, 1, \ldots, n$. It is useful because many questions ask only for a single term, such as the coefficient of a given power of $x$ or the constant (independent) term. You set the power of $x$ to the value you want, solve for $r$, and evaluate that one term, which is far quicker than expanding everything.

What numerator forms do you use for partial fractions?

The numerator form depends on the denominator factor. A distinct linear factor $(ax + b)$ gets a constant numerator $\dfrac{A}{ax + b}$. A repeated linear factor $(ax + b)^2$ needs two terms, $\dfrac{A}{ax + b} + \dfrac{B}{(ax + b)^2}$. An irreducible quadratic factor $(ax^2 + bx + c)$ gets a linear numerator $\dfrac{Ax + B}{ax^2 + bx + c}$. The fraction must be proper first; if not, divide out the polynomial part before decomposing.

How is this topic assessed in syllabus 4049?

Both Paper 1 and Paper 2 (each 2 hours 15 minutes, 90 marks, weighted 50 percent) draw on the whole syllabus, and candidates answer all questions. Binomial expansion questions often ask for a specific coefficient or the independent term, while partial fractions frequently appear as a step that prepares a fraction for later integration, so accurate decomposition matters beyond the immediate marks.

Does the binomial theorem in O-Level A-Maths cover negative or fractional powers?

No. Syllabus 4049 restricts the binomial theorem to expansions of $(a + b)^n$ where $n$ is a positive integer, so the expansion is finite with $n + 1$ terms. The infinite series for negative or fractional $n$ belongs to A-Level mathematics, not O-Level Additional Mathematics.

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Binomial theorem and partial fractions in O-Level Additional Mathematics (4049): expanding a power of a two-term expression, picking out a particular term, and splitting proper rational fractions

An O-Level Additional Mathematics (4049) overview of the binomial theorem and partial fractions in the Algebra strand. How to expand a plus b to a positive integer power, use the general term to find a specified coefficient or the constant term, and split a proper rational fraction into partial fractions, with links to every dot point.

Generated by Claude Opus 4.89 min readSEAB-4049 AlgebraUpdated 2026-06-13

Reviewed by: AI editorial process; not yet individually human-reviewed

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Two sides of the same coin
The binomial theorem
Finding a particular term
Partial fractions with linear factors
Repeated and quadratic factors
How this module is examined
Check your knowledge

Two sides of the same coin

This module pairs two Algebra-strand outcomes of Additional Mathematics (SEAB 4049) that are about handling structured expressions efficiently. The binomial theorem expands a power of a two-term expression without multiplying it out by hand, and partial fractions run the addition of fractions in reverse, breaking a single rational fraction into simpler pieces. Both reward a methodical, formula-driven approach rather than brute force.

The two outcomes also point forward. The general term of a binomial expansion is the standard tool for any "find the coefficient" question, and partial fractions are most often the preparation step that makes a fraction integrable in the Calculus strand. This overview ties the four dot points together; each has its own worked answers and practice.

The binomial theorem

The binomial theorem expands $(a + b)^n$ for a positive integer $n$ as a finite sum with $n + 1$ terms:

(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r},

where

\binom{n}{r} = \dfrac{n!}{r!(n-r)!}

are the binomial coefficients (the entries of Pascal's triangle). The powers of

a

count down while the powers of

b

count up, and the two powers always sum to

n

Finding a particular term

The particular term outcome uses the general term $T_{r+1} = \binom{n}{r} a^{n-r} b^{r}$ to extract one term without the full expansion. You write the power of $x$ that the general term produces, set it equal to the power you want (or to zero for the term independent of $x$ ), solve for $r$ , then evaluate that single term.

Find the term independent of

x

\left(2x - \dfrac{1}{x}\right)^6

step Write the general term

With $a = 2x$ , $b = -\dfrac{1}{x}$ and $n = 6$ :

T_{r+1} = \binom{6}{r}(2x)^{6-r}\left(-\frac{1}{x}\right)^{r} = \binom{6}{r} 2^{6-r}(-1)^{r}\, x^{6-r}\, x^{-r}.

step Collect the power of $x$

The power of $x$ is $(6 - r) - r = 6 - 2r$ . For the term independent of $x$ , set $6 - 2r = 0$ , so $r = 3$ .

step Evaluate that term

Substitute $r = 3$ :

T_4 = \binom{6}{3} 2^{3}(-1)^{3} = 20 \times 8 \times (-1) = -160.

The term independent of

x

-160

Partial fractions with linear factors

The partial fractions for proper fractions outcome splits a proper fraction whose denominator factorises into distinct linear factors. Each factor $(ax + b)$ contributes a term $\dfrac{A}{ax + b}$ , and you find the unknown constants either by substituting the value that makes a factor zero (the cover-up idea) or by comparing coefficients.

Repeated and quadratic factors

The repeated and quadratic factors outcome handles the harder denominator shapes. A repeated linear factor $(ax + b)^2$ needs both $\dfrac{A}{ax + b}$ and $\dfrac{B}{(ax + b)^2}$ , while an irreducible quadratic factor takes a linear numerator $\dfrac{Ax + B}{ax^2 + bx + c}$ . Always confirm the fraction is proper before decomposing; if the numerator degree is too high, divide first.

How this module is examined

Both papers, all questions. Paper 1 and Paper 2 (each 2 hours 15 minutes, 90 marks, 50 percent) cover the full syllabus, and you answer every question.
Read the term index carefully. $T_{r+1}$ means $r = 0$ gives the first term; mixing up $r$ and the term number is a common error in coefficient questions.
Match the numerator form to the factor. Using a constant numerator over a repeated or quadratic factor loses marks; the form must be correct before you solve for the constants.

Check your knowledge

Short questions on both outcomes. Work them with full method, then check the solutions.

Write the first three terms of the expansion of $(1 + 2x)^5$ in ascending powers of $x$ . (3 marks)
Find the coefficient of $x^2$ in the expansion of $(3 + x)^4$ . (2 marks)
Express $\dfrac{5x + 1}{(x + 1)(x - 2)}$ in partial fractions. (3 marks)
State the correct partial-fraction form (with unknown constants) for $\dfrac{7}{(x - 3)(x^2 + 1)}$ . (2 marks)

Solutions

These four questions cover the binomial expansion, finding a specific coefficient, decomposing into partial fractions, and stating the correct partial-fraction form. Work through each, then check your method below.

Step 1: Expand $(1 + 2x)^5$ up to $x^2$ in Q1

Use the binomial theorem with $a = 1$ , $b = 2x$ and $n = 5$ . The binomial coefficients for the first three terms are $\binom{5}{0} = 1$ , $\binom{5}{1} = 5$ and $\binom{5}{2} = 10$ . Substitute and simplify each power of $2x$ :

(1 + 2x)^5 = 1 + 5(2x) + 10(2x)^2 + \cdots = 1 + 10x + 10 \times 4x^2 + \cdots = 1 + 10x + 40x^2 + \cdots.

Final answer (Q1): $1 + 10x + 40x^2 + \cdots$

Step 2: Find the coefficient of $x^2$ in Q2

The general term of $(3 + x)^4$ is $\binom{4}{r} 3^{4-r} x^r$ . To find the $x^2$ term, set $r = 2$ and evaluate the coefficient. The $\binom{4}{r}$ factor counts the ways of choosing which copies of $x$ to take, and $3^{4-r}$ comes from the remaining factors:

\binom{4}{2} \cdot 3^{2} = 6 \times 9 = 54.

Final answer (Q2): coefficient of $x^2$ is $54$ .

Step 3: Decompose into partial fractions in Q3

Because the denominator has two distinct linear factors $(x + 1)$ and $(x - 2)$ , write the partial fractions with unknown constants $A$ and $B$ , then multiply through to clear the denominators:

\frac{5x + 1}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}, \qquad 5x + 1 = A(x - 2) + B(x + 1).

Substitute $x = -1$ to isolate $A$ : $-4 = -3A$ , so $A = \dfrac{4}{3}$ .
Substitute $x = 2$ to isolate $B$ : $11 = 3B$ , so $B = \dfrac{11}{3}$ .

\frac{5x + 1}{(x + 1)(x - 2)} = \frac{4}{3(x + 1)} + \frac{11}{3(x - 2)}.

Final answer (Q3): $\dfrac{4}{3(x + 1)} + \dfrac{11}{3(x - 2)}$ .

Step 4: State the correct partial-fraction form in Q4

A distinct linear factor in the denominator requires a constant numerator, while an irreducible quadratic factor (one that cannot be factorised over the reals) requires a linear numerator $Bx + C$ . Here $(x - 3)$ is a linear factor and $(x^2 + 1)$ is an irreducible quadratic, so the form is:

\frac{7}{(x - 3)(x^2 + 1)} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 1}.

Final answer (Q4): $\dfrac{A}{x - 3} + \dfrac{Bx + C}{x^2 + 1}$ .

Sources & how we know this

Singapore-Cambridge GCE O-Level Additional Mathematics (Syllabus 4049) — Singapore Examinations and Assessment Board (2026)