What is combining powers of x?

When both parts of the bracket contain x (for example x and 2x), use the index laws to combine them into a single power of x:

What is sign of a negative term?

With b = -1x, the factor (-1)^r alternates the sign; include it.

Find the term independent of x in (x - 1x)^4. [3 marks]

SEAB Original-style practice: Find the term independent of x in the expansion of (x + 2x)^6.

General term: 6rx^6 - r(2x)^r = 6r2^rx^6 - rx^-r = 6r2^rx^6 - 2r. The term independent of x has power zero: 6 - 2r = 0, so r = 3. Term: 632^3 = 20 × 8 = 160. Markers reward forming the general term, combining the powers of x, solving 6 - 2r = 0, and the value 160.

SingaporeAdditional MathematicsSyllabus dot point

How do we pick out a single term, such as the coefficient of x cubed or the constant term, from a binomial expansion?

Use the general term of a binomial expansion to find a specified term, the coefficient of a given power, or the term independent of x

A focused answer to the O-Level A-Maths outcome on the general term. Using the term formula to find a specified coefficient, a chosen power, or the constant term without writing the whole expansion.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

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

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What this dot point is asking
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What this dot point is asking

SEAB wants you to extract one specific term from a binomial expansion (a particular coefficient, the term in a chosen power, or the term that has no $x$ ) without writing out the whole expansion. The tool is the general term, which lets you jump straight to the term you need by solving for the right value of $r$ .

The answer

The general term

The term in $(a + b)^n$ that contains $b^r$ is:

T_{r+1} = \binom{n}{r}a^{n-r}b^r.

It is called the $(r+1)$ -th term because $r$ starts at $0$ . Knowing this single expression, you can target any term.

Finding the coefficient of a given power

To find the coefficient of $x^k$ , work out which power of $x$ the general term produces, set that equal to $k$ , and solve for $r$ . Then substitute $r$ back to evaluate the coefficient. The power of $x$ comes from both $a^{n-r}$ and $b^r$ if either contains $x$ .

Combining powers of x

When both parts of the bracket contain $x$ (for example $x$ and $\dfrac{2}{x}$ ), use the index laws to combine them into a single power of $x$ :

x^{n-r}\cdot x^{-r} = x^{n - 2r}.

The exponent then depends on $r$ , ready to be set equal to the power you want.

The term independent of x

The term independent of $x$ is the constant term, where the power of $x$ is zero. Set the combined exponent equal to $0$ , solve for $r$ , and evaluate that term.

When no such term exists

Solving for $r$ must give a whole number between $0$ and $n$ . If it gives a fraction or a value outside that range, the requested term simply does not appear in the expansion, and the correct answer is to say so rather than to force a term.

Coefficient versus term

Read the question carefully: the term in $x^3$ includes the $x^3$ , whereas the coefficient of $x^3$ is just the number in front. Likewise the term independent of $x$ is a pure number, which is both the term and its coefficient. Giving the wrong one of these is an easy mark to lose.

Worked example

Find the coefficient of $x^2$ in the expansion of $\left(2x - \dfrac{1}{x}\right)^4$ .

Step 1: Write the general term

T_{r+1} = \binom{4}{r}(2x)^{4-r}\left(-\frac{1}{x}\right)^{r} = \binom{4}{r}2^{4-r}(-1)^r x^{4-r}x^{-r}.

Step 2: Combine the powers of x

The power of $x$ is $(4 - r) + (-r) = 4 - 2r$ .

Step 3: Set the power equal to 2 and solve

$4 - 2r = 2 \Rightarrow r = 1$ .

Step 4: Evaluate the coefficient

\binom{4}{1}2^{3}(-1)^1 = 4 \times 8 \times (-1) = -32.

So the coefficient of $x^2$ is $-32$ .

Examples in context

Example 1. Picking a coefficient for a model. When a physical quantity is expanded as a power series, the coefficient of a particular power carries a specific physical meaning, and the general term lets an engineer read off just that coefficient without the full expansion.

Example 2. Constant background term. In a product expansion, the term independent of the variable represents the steady, variable-free part of a quantity; isolating it with the general term separates the constant contribution from the varying ones.

Try this

Q1. Write the general term of $(1 + x)^8$ . [1 mark]

Cue. $\binom{8}{r}x^{r}$ .

Q2. Find the coefficient of $x^2$ in $(1 + 3x)^5$ . [3 marks]

Cue. $\binom{5}{2}(3x)^2 = 10 \times 9 = 90$ , so the coefficient is $90$ .

Q3. Find the term independent of $x$ in $\left(x - \dfrac{1}{x}\right)^4$ . [3 marks]

Cue. Power $4 - 2r = 0$ gives $r = 2$ : $\binom{4}{2}(-1)^2 = 6$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksFind the coefficient of

x^3

in the expansion of

(2 + x)^6

Show worked answer →

The general term is $\binom{6}{r}2^{6 - r}x^{r}$ . For the $x^3$ term, set $r = 3$ .

$\binom{6}{3}2^{6 - 3}x^3 = 20 \times 2^3 \times x^3 = 20 \times 8 \times x^3 = 160x^3$ .

So the coefficient of $x^3$ is $160$ .

Markers reward the general term, choosing $r = 3$ , evaluating $\binom{6}{3} = 20$ and $2^3 = 8$ , and the coefficient $160$ .

Original5 marksFind the term independent of

x