The Binomial Expansion

What you'll learn

How to expand powers such as $(2+x2)6\left(2+\frac{x}{2}\right)^6$ without multiplying out six brackets.
How to write terms in ascending powers of $x$ .
How to use an expansion to estimate values like $1.04^8$ .
How to find unknown constants by comparing coefficients.

1. Brackets, terms, and coefficients

When you multiply out brackets, you are expanding them. For example, expanding $a+b)^3$ means multiplying $(a + b) (a + b) (a + b)$ and collecting like terms.

Definition

Key vocabulary

A binomial is an expression with two terms, such as $a + b$ or $2 - 5 x$ .
A term is one separate part of an expression, such as $12x^2$ .
The coefficient of a term is the number multiplying the variable part. In $12x^2$ , the coefficient of $x^2$ is 12.
Ascending powers of $x$ means writing terms in the order constant, $x$ , $x^2$ , $x^3$ , and so on.

A helpful way to remember the coefficients is Pascal’s triangle. Each row gives the coefficients for a power of a binomial.

Pascal’s triangle up to row 8, with row 5 highlighted to show the coefficients of (p+q)^5.

Pascal's triangle showing binomial coefficients up to row 8

Example

Fully expanding a fifth power

Expand $p+q)^5$ .

Use row 5 of Pascal’s triangle:
$11,\ 5,\ 10,\ 10,\ 5,\ 1$
The powers of $p$ go down from 5 to 0, while the powers of $q$ go up from 0 to 5.

The terms of (p+q)^5 showing the coefficient pattern and opposite movement of the powers of p and q.

Combine the coefficients with the powers:
$p+q)^5=p^5+5p^4q+10p^3q^2+10p^2q^3+5pq^4+q^5$

2. The binomial theorem

For AS-Level binomial expansion, you usually work with positive whole-number powers.

Key Idea

The finite binomial theorem

For a positive integer $n$ ,

(a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r

Here, $(nr)\binom{n}{r}$ is a binomial coefficient, read as “n choose r”.

You do not always need the full expansion. If the question asks for the first three terms in ascending powers of $x$ , you only need the terms with $x^0$ , $x^1$ , and $x^2$ .

Only the r=0, r=1, and r=2 terms are needed for the first three terms in ascending powers of x.

Example

First three terms

Find the first three terms in ascending powers of $x$ of $(2+x2)6\left(2+\frac{x}{2}\right)^6$ .

Identify the parts of the binomial:
$a=2,b=x2,n=6a=2,\qquad b=\frac{x}{2},\qquad n=6$
Use the terms for $r = 0$ , $r = 1$ , and $r = 2$ :
$r=0:(60)26=64r=1:(61)25(x2)=96xr=2:(62)24(x2)2=60x2\begin{aligned} r=0 &: \binom{6}{0}2^6=64\\ r=1 &: \binom{6}{1}2^5\left(\frac{x}{2}\right)=96x\\ r=2 &: \binom{6}{2}2^4\left(\frac{x}{2}\right)^2=60x^2 \end{aligned}$
Write the first three terms:
$(2+x2)6=64+96x+60x2+⋯\left(2+\frac{x}{2}\right)^6=64+96x+60x^2+\cdots$

Common Mistake

Forgetting the constant powers

In $(2+x2)6\left(2+\frac{x}{2}\right)^6$ , the $x$ term includes $2^5$ . Do not just do $6(x2)6\left(\frac{x}{2}\right)$ .

3. Negative signs and ascending powers

If the binomial contains a negative term, such as $2-5x)^7$ , the signs will alternate because odd powers of a negative expression are negative.

For (2-5x)^7, even powers of -5x give positive terms and odd powers give negative terms.

Example

A negative binomial

Find the first four terms in ascending powers of $x$ of $2-5x)^7$ .

Use the binomial theorem with $a = 2$ , $b = - 5 x$ , and $n = 7$ .
Work out the first four terms, using $r = 0$ to $r = 3$ :
$r=0:(70)27=128r=1:(71)26(−5x)=−2240xr=2:(72)25(−5x)2=16800x2r=3:(73)24(−5x)3=−70000x3\begin{aligned} r=0 &: \binom{7}{0}2^7=128\\ r=1 &: \binom{7}{1}2^6(-5x)=-2240x\\ r=2 &: \binom{7}{2}2^5(-5x)^2=16800x^2\\ r=3 &: \binom{7}{3}2^4(-5x)^3=-70000x^3 \end{aligned}$
Write the answer in ascending powers of $x$ :
$(2−5x)7=128−2240x+16800x2−70000x3+⋯(2-5x)^7=128-2240x+16800x^2-70000x^3+\cdots$

Tip

Sign check

If the $x$ part is negative, even powers give positive terms and odd powers give negative terms.

4. Using an expansion as an approximation

You can use the first few terms of an expansion to estimate powers of numbers close to a convenient value.

For example, $1.04^8$ is close to $1^8$ , so an expansion of $1+4x)^8$ is useful because $1 + 4 x = 1.04$ when $x = 0.01$ .

Matching 1+4x to 1.04 gives a small value x=0.01, making the truncated expansion useful.

Example

Approximating a power

Use the expansion of $1+4x)^8$ up to the $x^3$ term to estimate $1.04^8$ .

Expand up to and including the $x^3$ term:
$(1+4x)8=1+8(4x)+(82)(4x)2+(83)(4x)3+⋯=1+32x+448x2+3584x3+⋯\begin{aligned} (1+4x)^8 &=1+8(4x)+\binom{8}{2}(4x)^2+\binom{8}{3}(4x)^3+\cdots\\ &=1+32x+448x^2+3584x^3+\cdots \end{aligned}$
Match the bracket to $1.04$ :
$1 + 4 x = 1.04$
Solve for $x$ :
$x = 0.01$
Substitute $x = 0.01$ into the truncated expansion:
$1+32(0.01)+448(0.01)^2+3584(0.01)^3=1.368384$
Give the estimate to 5 significant figures:
$1.048≈1.36841.04^8\approx 1.3684$

Tip

Match before substituting

Always solve the inside of the bracket first. For example, if you have $(3+2x5)6\left(3+\frac{2x}{5}\right)^6$ and want $2.92^6$ , solve $3+2x5=2.923+\frac{2x}{5}=2.92$ .

5. Multiplying by another factor

Sometimes you expand a binomial and then multiply it by something like $a x + b$ . If you only need the first two terms, do not expand everything.

Example

Finding a and b from first terms

Let $f(x)=(ax+b)(2+x8)8f(x)=(ax+b)\left(2+\frac{x}{8}\right)^8$ . Given that the first two terms of $f (x)$ are $40 + 72 x$ , find $a$ and $b$ .

First find only the constant and $x$ terms of the binomial:
$(2+x8)8=256+128x+⋯\left(2+\frac{x}{8}\right)^8=256+128x+\cdots$
Multiply by $a x + b$ and focus on the constant and $x$ terms:

A product grid shows which combinations contribute to the constant term and the coefficient of x.

$$
(ax+b)(256+128x+\cdots)
$$

3. The constant term comes from $b⋅256b\cdot 256$ :

$$
256b=40
$$

4. Solve for $b$ :

$$
b=\frac{5}{32}
$$

5. The coefficient of $x$ comes from $ax⋅256ax\cdot 256$ and $b⋅128xb\cdot 128x$ :

$$
256a+128b=72
$$

6. Substitute $b=532b=\frac{5}{32}$ :

$$
256a+20=72
$$

7. Solve for $a$ :

$$
a=\frac{13}{64}
$$

Common Mistake

Using too many terms

For the first two terms of $(ax+b)(binomial)(ax+b)(\text{binomial})$ , you usually only need the constant and $x$ terms of the binomial.

6. Finding unknown constants from coefficients

A common exam move is to compare coefficients. This means matching the coefficient of a particular power of $x$ to the information given.

Example

Solving for k

In the expansion of $1+kx)^{10}$ , the coefficient of $x^3$ is twice the coefficient of $x$ . Given that $k≠0k\neq 0$ , find $k$ .

Write the first few terms:
$(1+kx)10=1+10kx+45k2x2+120k3x3+⋯(1+kx)^{10}=1+10kx+45k^2x^2+120k^3x^3+\cdots$
Identify the two coefficients:
$x3=120k3\text{coefficient of }x=10k,\qquad \text{coefficient of }x^3=120k^3$
Use the condition “the coefficient of $x^3$ is twice the coefficient of $x$ ”:
$120k^3=2(10k)$
Since $k≠0k\neq 0$ , divide by $k$ :
$120k^2=20$
Solve for $k$ :
$k=±16k=\pm\frac{1}{\sqrt{6}}$

7. Probability and binomial expansion

Binomial expansion is also linked to repeated independent events with two outcomes, such as late/not late or success/failure.

If $p$ is the probability of one outcome and $q$ is the probability of the other, then $p+q)^5$ lists the probabilities for 5 repeated trials, with coefficients counting the number of arrangements.

The coefficients in (p+q)^5 count how many arrangements have each number of successes.

Example

No more than once

A student has probability 0.2 of being late on any school day. Find the probability that they are late no more than once in a 5-day week.

Let $p = 0.2$ be the probability of being late and $q = 0.8$ be the probability of not being late.
“No more than once” means 0 times late or 1 time late.
Use the relevant terms from $p+q)^5$ :
$q^5+5pq^4$
Substitute $p = 0.2$ and $q = 0.8$ :
$0.8^5+5(0.2)(0.8)^4=0.73728$

8. Proof-style questions

Sometimes you expand expressions to show something is always true. Look for positive even powers such as $x^2$ and $x^4$ .

Adding (3+x)^4 and (3-x)^4 cancels the odd-power terms and leaves only non-negative even-power terms plus 162.

Example

Using cancellation to prove an inequality

Prove that $(3+x)4+(3−x)4≥162(3+x)^4+(3-x)^4\ge 162$ .

Expand both expressions:
$(3+x)4=81+108x+54x2+12x3+x4(3−x)4=81−108x+54x2−12x3+x4\begin{aligned} (3+x)^4&=81+108x+54x^2+12x^3+x^4\\ (3-x)^4&=81-108x+54x^2-12x^3+x^4 \end{aligned}$
Add them together:
$3+x)^4+(3-x)^4=162+108x^2+2x^4$
Since $x2≥0x^2\ge 0$ and $x4≥0x^4\ge 0$ , the extra part is never negative:
$108x2+2x4≥0108x^2+2x^4\ge 0$
Therefore:
$(3+x)4+(3−x)4≥162(3+x)^4+(3-x)^4\ge 162$

Exam technique

In the exam

Identify the binomial parts carefully: the constant part, the $x$ part, and the power.
If the question asks for “first three terms”, only use $r = 0$ , $r = 1$ , and $r = 2$ .
For coefficient questions, write the relevant terms clearly, then equate coefficients of the same power of $x$ .

Self review

Check yourself

What are the first three terms of $1-3x)^6$ in ascending powers of $x$ ?
How would you choose $x$ to use an expansion of $(2+x4)5\left(2+\frac{x}{4}\right)^5$ to estimate $2.01^5$ ?
In $1+kx)^8$ , which term contains the coefficient of $x^3$ ?

The Binomial Expansion

What you'll learn

How to expand powers such as $(2+x2)6\left(2+\frac{x}{2}\right)^6$ without multiplying out six brackets.
How to write terms in ascending powers of $x$ .
How to use an expansion to estimate values like $1.04^8$ .
How to find unknown constants by comparing coefficients.

1. Brackets, terms, and coefficients

When you multiply out brackets, you are expanding them. For example, expanding $a+b)^3$ means multiplying $(a + b) (a + b) (a + b)$ and collecting like terms.

Definition

Key vocabulary

A binomial is an expression with two terms, such as $a + b$ or $2 - 5 x$ .
A term is one separate part of an expression, such as $12x^2$ .
The coefficient of a term is the number multiplying the variable part. In $12x^2$ , the coefficient of $x^2$ is 12.
Ascending powers of $x$ means writing terms in the order constant, $x$ , $x^2$ , $x^3$ , and so on.

A helpful way to remember the coefficients is Pascal’s triangle. Each row gives the coefficients for a power of a binomial.

Pascal’s triangle up to row 8, with row 5 highlighted to show the coefficients of (p+q)^5.

Pascal's triangle showing binomial coefficients up to row 8

Example

Fully expanding a fifth power

Expand $p+q)^5$ .

Use row 5 of Pascal’s triangle:
$11,\ 5,\ 10,\ 10,\ 5,\ 1$
The powers of $p$ go down from 5 to 0, while the powers of $q$ go up from 0 to 5.

The terms of (p+q)^5 showing the coefficient pattern and opposite movement of the powers of p and q.

Combine the coefficients with the powers:
$p+q)^5=p^5+5p^4q+10p^3q^2+10p^2q^3+5pq^4+q^5$

2. The binomial theorem

For AS-Level binomial expansion, you usually work with positive whole-number powers.

Key Idea

The finite binomial theorem

For a positive integer $n$ ,

(a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r

Here, $(nr)\binom{n}{r}$ is a binomial coefficient, read as “n choose r”.

You do not always need the full expansion. If the question asks for the first three terms in ascending powers of $x$ , you only need the terms with $x^0$ , $x^1$ , and $x^2$ .

Only the r=0, r=1, and r=2 terms are needed for the first three terms in ascending powers of x.

Example

First three terms

Find the first three terms in ascending powers of $x$ of $(2+x2)6\left(2+\frac{x}{2}\right)^6$ .

Identify the parts of the binomial:
$a=2,b=x2,n=6a=2,\qquad b=\frac{x}{2},\qquad n=6$
Use the terms for $r = 0$ , $r = 1$ , and $r = 2$ :
$r=0:(60)26=64r=1:(61)25(x2)=96xr=2:(62)24(x2)2=60x2\begin{aligned} r=0 &: \binom{6}{0}2^6=64\\ r=1 &: \binom{6}{1}2^5\left(\frac{x}{2}\right)=96x\\ r=2 &: \binom{6}{2}2^4\left(\frac{x}{2}\right)^2=60x^2 \end{aligned}$
Write the first three terms:
$(2+x2)6=64+96x+60x2+⋯\left(2+\frac{x}{2}\right)^6=64+96x+60x^2+\cdots$

Common Mistake

Forgetting the constant powers

In $(2+x2)6\left(2+\frac{x}{2}\right)^6$ , the $x$ term includes $2^5$ . Do not just do $6(x2)6\left(\frac{x}{2}\right)$ .

3. Negative signs and ascending powers

If the binomial contains a negative term, such as $2-5x)^7$ , the signs will alternate because odd powers of a negative expression are negative.

For (2-5x)^7, even powers of -5x give positive terms and odd powers give negative terms.

Example

A negative binomial

Find the first four terms in ascending powers of $x$ of $2-5x)^7$ .

Use the binomial theorem with $a = 2$ , $b = - 5 x$ , and $n = 7$ .
Work out the first four terms, using $r = 0$ to $r = 3$ :
$r=0:(70)27=128r=1:(71)26(−5x)=−2240xr=2:(72)25(−5x)2=16800x2r=3:(73)24(−5x)3=−70000x3\begin{aligned} r=0 &: \binom{7}{0}2^7=128\\ r=1 &: \binom{7}{1}2^6(-5x)=-2240x\\ r=2 &: \binom{7}{2}2^5(-5x)^2=16800x^2\\ r=3 &: \binom{7}{3}2^4(-5x)^3=-70000x^3 \end{aligned}$
Write the answer in ascending powers of $x$ :
$(2−5x)7=128−2240x+16800x2−70000x3+⋯(2-5x)^7=128-2240x+16800x^2-70000x^3+\cdots$

Tip

Sign check

If the $x$ part is negative, even powers give positive terms and odd powers give negative terms.

4. Using an expansion as an approximation

You can use the first few terms of an expansion to estimate powers of numbers close to a convenient value.

For example, $1.04^8$ is close to $1^8$ , so an expansion of $1+4x)^8$ is useful because $1 + 4 x = 1.04$ when $x = 0.01$ .

Matching 1+4x to 1.04 gives a small value x=0.01, making the truncated expansion useful.

Example

Approximating a power

Use the expansion of $1+4x)^8$ up to the $x^3$ term to estimate $1.04^8$ .

Expand up to and including the $x^3$ term:
$(1+4x)8=1+8(4x)+(82)(4x)2+(83)(4x)3+⋯=1+32x+448x2+3584x3+⋯\begin{aligned} (1+4x)^8 &=1+8(4x)+\binom{8}{2}(4x)^2+\binom{8}{3}(4x)^3+\cdots\\ &=1+32x+448x^2+3584x^3+\cdots \end{aligned}$
Match the bracket to $1.04$ :
$1 + 4 x = 1.04$
Solve for $x$ :
$x = 0.01$
Substitute $x = 0.01$ into the truncated expansion:
$1+32(0.01)+448(0.01)^2+3584(0.01)^3=1.368384$
Give the estimate to 5 significant figures:
$1.048≈1.36841.04^8\approx 1.3684$

Tip

Match before substituting

Always solve the inside of the bracket first. For example, if you have $(3+2x5)6\left(3+\frac{2x}{5}\right)^6$ and want $2.92^6$ , solve $3+2x5=2.923+\frac{2x}{5}=2.92$ .

5. Multiplying by another factor

Sometimes you expand a binomial and then multiply it by something like $a x + b$ . If you only need the first two terms, do not expand everything.

Example

Finding a and b from first terms

Let $f(x)=(ax+b)(2+x8)8f(x)=(ax+b)\left(2+\frac{x}{8}\right)^8$ . Given that the first two terms of $f (x)$ are $40 + 72 x$ , find $a$ and $b$ .

First find only the constant and $x$ terms of the binomial:
$(2+x8)8=256+128x+⋯\left(2+\frac{x}{8}\right)^8=256+128x+\cdots$
Multiply by $a x + b$ and focus on the constant and $x$ terms:

A product grid shows which combinations contribute to the constant term and the coefficient of x.

$$
(ax+b)(256+128x+\cdots)
$$

3. The constant term comes from $b⋅256b\cdot 256$ :

$$
256b=40
$$

4. Solve for $b$ :

$$
b=\frac{5}{32}
$$

5. The coefficient of $x$ comes from $ax⋅256ax\cdot 256$ and $b⋅128xb\cdot 128x$ :

$$
256a+128b=72
$$

6. Substitute $b=532b=\frac{5}{32}$ :

$$
256a+20=72
$$

7. Solve for $a$ :

$$
a=\frac{13}{64}
$$

Common Mistake

Using too many terms

For the first two terms of $(ax+b)(binomial)(ax+b)(\text{binomial})$ , you usually only need the constant and $x$ terms of the binomial.

6. Finding unknown constants from coefficients

A common exam move is to compare coefficients. This means matching the coefficient of a particular power of $x$ to the information given.

Example

Solving for k

In the expansion of $1+kx)^{10}$ , the coefficient of $x^3$ is twice the coefficient of $x$ . Given that $k≠0k\neq 0$ , find $k$ .

Write the first few terms:
$(1+kx)10=1+10kx+45k2x2+120k3x3+⋯(1+kx)^{10}=1+10kx+45k^2x^2+120k^3x^3+\cdots$
Identify the two coefficients:
$x3=120k3\text{coefficient of }x=10k,\qquad \text{coefficient of }x^3=120k^3$
Use the condition “the coefficient of $x^3$ is twice the coefficient of $x$ ”:
$120k^3=2(10k)$
Since $k≠0k\neq 0$ , divide by $k$ :
$120k^2=20$
Solve for $k$ :
$k=±16k=\pm\frac{1}{\sqrt{6}}$

7. Probability and binomial expansion

Binomial expansion is also linked to repeated independent events with two outcomes, such as late/not late or success/failure.

If $p$ is the probability of one outcome and $q$ is the probability of the other, then $p+q)^5$ lists the probabilities for 5 repeated trials, with coefficients counting the number of arrangements.

The coefficients in (p+q)^5 count how many arrangements have each number of successes.

Example

No more than once

A student has probability 0.2 of being late on any school day. Find the probability that they are late no more than once in a 5-day week.

Let $p = 0.2$ be the probability of being late and $q = 0.8$ be the probability of not being late.
“No more than once” means 0 times late or 1 time late.
Use the relevant terms from $p+q)^5$ :
$q^5+5pq^4$
Substitute $p = 0.2$ and $q = 0.8$ :
$0.8^5+5(0.2)(0.8)^4=0.73728$

8. Proof-style questions

Sometimes you expand expressions to show something is always true. Look for positive even powers such as $x^2$ and $x^4$ .

Adding (3+x)^4 and (3-x)^4 cancels the odd-power terms and leaves only non-negative even-power terms plus 162.

Example

Using cancellation to prove an inequality

Prove that $(3+x)4+(3−x)4≥162(3+x)^4+(3-x)^4\ge 162$ .

Expand both expressions:
$(3+x)4=81+108x+54x2+12x3+x4(3−x)4=81−108x+54x2−12x3+x4\begin{aligned} (3+x)^4&=81+108x+54x^2+12x^3+x^4\\ (3-x)^4&=81-108x+54x^2-12x^3+x^4 \end{aligned}$
Add them together:
$3+x)^4+(3-x)^4=162+108x^2+2x^4$
Since $x2≥0x^2\ge 0$ and $x4≥0x^4\ge 0$ , the extra part is never negative:
$108x2+2x4≥0108x^2+2x^4\ge 0$
Therefore:
$(3+x)4+(3−x)4≥162(3+x)^4+(3-x)^4\ge 162$

Exam technique

In the exam

Identify the binomial parts carefully: the constant part, the $x$ part, and the power.
If the question asks for “first three terms”, only use $r = 0$ , $r = 1$ , and $r = 2$ .
For coefficient questions, write the relevant terms clearly, then equate coefficients of the same power of $x$ .

Self review

Check yourself

What are the first three terms of $1-3x)^6$ in ascending powers of $x$ ?
How would you choose $x$ to use an expansion of $(2+x4)5\left(2+\frac{x}{4}\right)^5$ to estimate $2.01^5$ ?
In $1+kx)^8$ , which term contains the coefficient of $x^3$ ?

The Binomial Expansion

What you'll learn

1. Brackets, terms, and coefficients

2. The binomial theorem

3. Negative signs and ascending powers

4. Using an expansion as an approximation

5. Multiplying by another factor

6. Finding unknown constants from coefficients

7. Probability and binomial expansion

8. Proof-style questions

Recap questions

You've reached the end

The Binomial Expansion

What you'll learn

1. Brackets, terms, and coefficients

2. The binomial theorem

3. Negative signs and ascending powers

4. Using an expansion as an approximation

5. Multiplying by another factor

6. Finding unknown constants from coefficients

7. Probability and binomial expansion

8. Proof-style questions

Recap questions

You've reached the end

The Binomial Expansion

What you'll learn

1. Brackets, terms, and coefficients

2. The binomial theorem

3. Negative signs and ascending powers

4. Using an expansion as an approximation

5. Multiplying by another factor

6. Finding unknown constants from coefficients

7. Probability and binomial expansion

8. Proof-style questions

Recap questions

You've reached the end

The Binomial Expansion Revision Guide

The Binomial Expansion

What you'll learn

1. Brackets, terms, and coefficients

2. The binomial theorem

3. Negative signs and ascending powers

4. Using an expansion as an approximation

5. Multiplying by another factor

6. Finding unknown constants from coefficients

7. Probability and binomial expansion

8. Proof-style questions

Recap questions

You've reached the end

The Binomial Expansion Revision Guide