Binomial Expansion

What you'll learn

Expand expressions like $(1+3x)−12(1+3x)^{-\frac{1}{2}}$ in ascending powers of $x$ .
Factor expressions such as $(9+6x)−12(9+6x)^{-\frac{1}{2}}$ into the standard binomial form.
State the range of values for which an expansion is valid.
Use expansions for approximations and coefficient-matching questions.

1. The language of binomial expansion

A binomial is an expression with two terms, such as $1 + x$ or $4 + 6 x$ . A binomial expansion rewrites a power of a binomial as a series, meaning a sum of terms.

You will often need index laws first:

$1+6x=(1+6x)12\sqrt{1+6x}=(1+6x)^{\frac{1}{2}}$
$19+x=(9+x)−12\frac{1}{\sqrt{9+x}}=(9+x)^{-\frac{1}{2}}$
$1(2+3x)2=(2+3x)−2\frac{1}{(2+3x)^2}=(2+3x)^{-2}$

Definition

Key vocabulary

A term is one part of a sum, such as $6x^2$ .
A coefficient is the number multiplying a power of the variable, so the coefficient of $x^2$ in $6x^2$ is 6.
Ascending powers of $x$ means constant term first, then $x$ , then $x^2$ , then $x^3$ , and so on.

2. Positive integer powers

If the power is a non-negative integer, meaning 0, 1, 2, 3, and so on, the expansion is finite: it stops.

For a positive integer power,

A schematic showing how each term in a finite binomial expansion is built from the choice of powers of the two parts of the bracket.

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

The symbol $(nr)\binom{n}{r}$ is called a binomial coefficient, read “ $n$ choose $r$ ”. It is given by

(nr)=n!r!(n−r)!\binom{n}{r}=\frac{n!}{r!(n-r)!}

where $n!$ means $n⋅(n−1)⋅(n−2)⋯1n\cdot(n-1)\cdot(n-2)\cdots 1$ .

Example

Finding a coefficient

Find the coefficient of $x^5$ in the expansion of $3+2x)^7$ .

The general term comes from choosing $r$ copies of $2 x$ and $7 - r$ copies of 3.
To get $x^5$ , use $r = 5$ .
Substitute into the binomial term:
$(75)37−5(2x)5\binom{7}{5}3^{7-5}(2x)^5$
Simplify the coefficient:
$(75)32⋅25=21⋅9⋅32=6048\binom{7}{5}3^2\cdot 2^5=21\cdot 9\cdot 32=6048$
So the coefficient of $x^5$ is 6048.

3. Fractional and negative powers

For powers such as $−12-\frac{1}{2}$ , $13\frac{1}{3}$ , or $- 3$ , the expansion usually goes on forever. These are handled using the general binomial expansion:

(1+u)n=1+nu+n(n−1)2u2+n(n−1)(n−2)6u3+⋯(1+u)^n=1+nu+\frac{n(n-1)}{2}u^2+\frac{n(n-1)(n-2)}{6}u^3+\cdots

Here, $u$ is the expression inside the bracket after the 1, and $n$ is the power.

The standard binomial form highlights the roles of 1, u and n in an expression like (1+u)^n.

Key Idea

The standard form

The infinite binomial formula is built for $1+u)^n$ . If your bracket is not in that form, factor it first.

Example

Expanding a negative power

Expand $1-2x)^{-3}$ up to and including the term in $x^3$ , and state the range of validity.

Identify the two key parts: $n = - 3$ and $u = - 2 x$ .
Substitute into the first four terms of the formula:
$(1−2x)−3=1+(−3)(−2x)+(−3)(−4)2(−2x)2+(−3)(−4)(−5)6(−2x)3+⋯(1-2x)^{-3}=1+(-3)(-2x)+\frac{(-3)(-4)}{2}(-2x)^2+\frac{(-3)(-4)(-5)}{6}(-2x)^3+\cdots$
Simplify each term:
$(1−2x)−3=1+6x+24x2+80x3+⋯(1-2x)^{-3}=1+6x+24x^2+80x^3+\cdots$
For a negative or fractional power, require $∣ u ∣ < 1$ :

The validity condition |u|<1 gives an open interval for x after solving the inequality.

$$
|-2x|<1\Rightarrow |x|<\frac{1}{2}\Rightarrow -\frac{1}{2}<x<\frac{1}{2}
$$

Common Mistake

Miscounting the terms

“Up to and including $x^3$ ” means include the constant term, the $x$ term, the $x^2$ term, and the $x^3$ term. Do not stop at $x^2$ .

4. Factoring first

Many expressions do not start as $1+u)^n$ . For example, $(9+6x)−12(9+6x)^{-\frac{1}{2}}$ has a 9 at the front, not a 1.

Use this idea:

Factoring out the constant turns a bracket into the standard binomial form needed for the infinite expansion.

(a+bx)n=an(1+bax)n(a+bx)^n=a^n\left(1+\frac{b}{a}x\right)^n

Then use the binomial expansion on the bracket.

Example

Factoring into standard form

Expand $(9+6x)−12(9+6x)^{-\frac{1}{2}}$ up to and including the term in $x^3$ , and state the range of validity.

Factor 9 from the bracket:
$(9+6x)−12=9−12(1+23x)−12=13(1+23x)−12(9+6x)^{-\frac{1}{2}}=9^{-\frac{1}{2}}\left(1+\frac{2}{3}x\right)^{-\frac{1}{2}}=\frac{1}{3}\left(1+\frac{2}{3}x\right)^{-\frac{1}{2}}$
Expand the bracket using $n=−12n=-\frac{1}{2}$ and $u=23xu=\frac{2}{3}x$ :
$(1+23x)−12=1−x3+x26−5x354+⋯\left(1+\frac{2}{3}x\right)^{-\frac{1}{2}}=1-\frac{x}{3}+\frac{x^2}{6}-\frac{5x^3}{54}+\cdots$
Multiply every term by $13\frac{1}{3}$ :
$(9+6x)−12=13−x9+x218−5x3162+⋯(9+6x)^{-\frac{1}{2}}=\frac{1}{3}-\frac{x}{9}+\frac{x^2}{18}-\frac{5x^3}{162}+\cdots$
The range comes from $∣23x∣<1\left|\frac{2}{3}x\right|<1$ :
$−32<x<32-\frac{3}{2}<x<\frac{3}{2}$

Common Mistake

Expanding the wrong bracket

Do not apply the formula directly to $9 + 6 x$ as though it were $1 + u$ . The constant factor must come out first.

5. Validity and accuracy

For fractional or negative powers, the expansion is valid when

∣ u ∣ < 1

where $u$ is the small expression in $1+u)^n$ .

For example, if you have $(1−2x9)12\left(1-\frac{2x}{9}\right)^{\frac{1}{2}}$ , then $u=−2x9u=-\frac{2x}{9}$ , so the condition is $∣−2x9∣<1\left|-\frac{2x}{9}\right|<1$ .

This number line shows the range of x values for which the expansion of (1−2x/9)^(1/2) is valid.

Common Mistake

Validity before accuracy

If $∣u∣≥1|u|\ge 1$ , the expansion should not be used. Even when it is valid, it is usually more accurate when $u$ is close to 0.

6. Using expansions for approximations

A binomial expansion is especially useful when the value of $x$ is small. Higher powers like $x^2$ and $x^3$ become much smaller, so the first few terms give a good approximation.

For small x, successive powers shrink rapidly, so later terms usually contribute much less to the approximation.

Example

Approximating a square root

Use a binomial expansion to estimate $8.8\sqrt{8.8}$ to 5 decimal places.

Write $8.8\sqrt{8.8}$ as part of $(9−2x)12(9-2x)^{\frac{1}{2}}$ :
$9 - 2 x = 8.8$
Solve for the substitution value:
$x = 0.1$
Expand $(9−2x)12(9-2x)^{\frac{1}{2}}$ :
$(9−2x)12=3(1−2x9)12=3−x3−x254−x3486+⋯\begin{aligned} (9-2x)^{\frac{1}{2}} &=3\left(1-\frac{2x}{9}\right)^{\frac{1}{2}}\\ &=3-\frac{x}{3}-\frac{x^2}{54}-\frac{x^3}{486}+\cdots \end{aligned}$
Substitute $x = 0.1$ :
$3−0.13−0.1254−0.13486=2.966479…3-\frac{0.1}{3}-\frac{0.1^2}{54}-\frac{0.1^3}{486}=2.966479\ldots$
Therefore, $8.8≈2.96648\sqrt{8.8}\approx 2.96648$ to 5 decimal places.

7. Combining expansions with other algebra

Sometimes you expand one part first, then multiply by another factor or match coefficients.

Multiplying by another factor

Example

Multiplying after expanding

Find the expansion of $1+x(1−2x)2\frac{1+x}{(1-2x)^2}$ up to and including $x^3$ .

First expand the denominator part:
$(1−2x)−2=1+4x+12x2+32x3+⋯(1-2x)^{-2}=1+4x+12x^2+32x^3+\cdots$
Multiply by $1 + x$ , keeping terms only up to $x^3$ :

A multiplication grid makes clear which products contribute up to x^3 and which are ignored.

$$
(1+x)(1+4x+12x^2+32x^3)=1+5x+16x^2+44x^3+\cdots
$$

3. Terms like $x⋅32x3x\cdot 32x^3$ are in $x^4$ , so they are not needed.

Tip

Only keep what can matter

When multiplying expansions up to $x^3$ , ignore any product whose powers already add to more than 3.

Matching coefficients

Example

Finding unknown constants

The expansion of $a+bx4+x\frac{a+bx}{\sqrt{4+x}}$ begins $2+x+⋯2+x+\cdots$ . Find $a$ and $b$ .

Begin with the reciprocal square root expansion:
$(4+x)−12=12−x16+⋯(4+x)^{-\frac{1}{2}}=\frac{1}{2}-\frac{x}{16}+\cdots$
Multiply by $a + b x$ and keep the constant and $x$ terms:
$)=a2+(b2−a16)x+⋯(a+bx)\left(\frac{1}{2}-\frac{x}{16}+\cdots\right)=\frac{a}{2}+\left(\frac{b}{2}-\frac{a}{16}\right)x+\cdots$
Compare coefficients with $2+x+⋯2+x+\cdots$ :
$a2=2,b2−a16=1\frac{a}{2}=2,\qquad \frac{b}{2}-\frac{a}{16}=1$
Solve the equations: $a = 4$ , then $b2−14=1\frac{b}{2}-\frac{1}{4}=1$ , so $b=52b=\frac{5}{2}$ .

Exam technique

In the exam

Rewrite roots and fractions as powers before expanding.
Factor the bracket into $A(1+u)^n$ if the constant term is not 1.
For fractional or negative powers, always state the range using $∣ u ∣ < 1$ .
In approximation questions, choose the substitution with the smallest valid value of $∣ u ∣$ .
When multiplying expansions, keep only the terms that can contribute to the requested power.

Self review

Check yourself

Can you expand $(1−3x)−12(1-3x)^{-\frac{1}{2}}$ up to and including $x^2$ , and state its validity range?
Can you rewrite $125+10x\frac{1}{\sqrt{25+10x}}$ in the form $A(1+u)^n$ ?
If you multiply an expansion by $1 + 2 x$ , which terms can contribute to the coefficient of $x^3$ ?

Binomial Expansion

What you'll learn

Expand expressions like $(1+3x)−12(1+3x)^{-\frac{1}{2}}$ in ascending powers of $x$ .
Factor expressions such as $(9+6x)−12(9+6x)^{-\frac{1}{2}}$ into the standard binomial form.
State the range of values for which an expansion is valid.
Use expansions for approximations and coefficient-matching questions.

1. The language of binomial expansion

A binomial is an expression with two terms, such as $1 + x$ or $4 + 6 x$ . A binomial expansion rewrites a power of a binomial as a series, meaning a sum of terms.

You will often need index laws first:

$1+6x=(1+6x)12\sqrt{1+6x}=(1+6x)^{\frac{1}{2}}$
$19+x=(9+x)−12\frac{1}{\sqrt{9+x}}=(9+x)^{-\frac{1}{2}}$
$1(2+3x)2=(2+3x)−2\frac{1}{(2+3x)^2}=(2+3x)^{-2}$

Definition

Key vocabulary

A term is one part of a sum, such as $6x^2$ .
A coefficient is the number multiplying a power of the variable, so the coefficient of $x^2$ in $6x^2$ is 6.
Ascending powers of $x$ means constant term first, then $x$ , then $x^2$ , then $x^3$ , and so on.

2. Positive integer powers

If the power is a non-negative integer, meaning 0, 1, 2, 3, and so on, the expansion is finite: it stops.

For a positive integer power,

A schematic showing how each term in a finite binomial expansion is built from the choice of powers of the two parts of the bracket.

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

The symbol $(nr)\binom{n}{r}$ is called a binomial coefficient, read “ $n$ choose $r$ ”. It is given by

(nr)=n!r!(n−r)!\binom{n}{r}=\frac{n!}{r!(n-r)!}

where $n!$ means $n⋅(n−1)⋅(n−2)⋯1n\cdot(n-1)\cdot(n-2)\cdots 1$ .

Example

Finding a coefficient

Find the coefficient of $x^5$ in the expansion of $3+2x)^7$ .

The general term comes from choosing $r$ copies of $2 x$ and $7 - r$ copies of 3.
To get $x^5$ , use $r = 5$ .
Substitute into the binomial term:
$(75)37−5(2x)5\binom{7}{5}3^{7-5}(2x)^5$
Simplify the coefficient:
$(75)32⋅25=21⋅9⋅32=6048\binom{7}{5}3^2\cdot 2^5=21\cdot 9\cdot 32=6048$
So the coefficient of $x^5$ is 6048.

3. Fractional and negative powers

For powers such as $−12-\frac{1}{2}$ , $13\frac{1}{3}$ , or $- 3$ , the expansion usually goes on forever. These are handled using the general binomial expansion:

(1+u)n=1+nu+n(n−1)2u2+n(n−1)(n−2)6u3+⋯(1+u)^n=1+nu+\frac{n(n-1)}{2}u^2+\frac{n(n-1)(n-2)}{6}u^3+\cdots

Here, $u$ is the expression inside the bracket after the 1, and $n$ is the power.

The standard binomial form highlights the roles of 1, u and n in an expression like (1+u)^n.

Key Idea

The standard form

The infinite binomial formula is built for $1+u)^n$ . If your bracket is not in that form, factor it first.

Example

Expanding a negative power

Expand $1-2x)^{-3}$ up to and including the term in $x^3$ , and state the range of validity.

Identify the two key parts: $n = - 3$ and $u = - 2 x$ .
Substitute into the first four terms of the formula:
$(1−2x)−3=1+(−3)(−2x)+(−3)(−4)2(−2x)2+(−3)(−4)(−5)6(−2x)3+⋯(1-2x)^{-3}=1+(-3)(-2x)+\frac{(-3)(-4)}{2}(-2x)^2+\frac{(-3)(-4)(-5)}{6}(-2x)^3+\cdots$
Simplify each term:
$(1−2x)−3=1+6x+24x2+80x3+⋯(1-2x)^{-3}=1+6x+24x^2+80x^3+\cdots$
For a negative or fractional power, require $∣ u ∣ < 1$ :

The validity condition |u|<1 gives an open interval for x after solving the inequality.

$$
|-2x|<1\Rightarrow |x|<\frac{1}{2}\Rightarrow -\frac{1}{2}<x<\frac{1}{2}
$$

Common Mistake

Miscounting the terms

“Up to and including $x^3$ ” means include the constant term, the $x$ term, the $x^2$ term, and the $x^3$ term. Do not stop at $x^2$ .

4. Factoring first

Many expressions do not start as $1+u)^n$ . For example, $(9+6x)−12(9+6x)^{-\frac{1}{2}}$ has a 9 at the front, not a 1.

Use this idea:

Factoring out the constant turns a bracket into the standard binomial form needed for the infinite expansion.

(a+bx)n=an(1+bax)n(a+bx)^n=a^n\left(1+\frac{b}{a}x\right)^n

Then use the binomial expansion on the bracket.

Example

Factoring into standard form

Expand $(9+6x)−12(9+6x)^{-\frac{1}{2}}$ up to and including the term in $x^3$ , and state the range of validity.

Factor 9 from the bracket:
$(9+6x)−12=9−12(1+23x)−12=13(1+23x)−12(9+6x)^{-\frac{1}{2}}=9^{-\frac{1}{2}}\left(1+\frac{2}{3}x\right)^{-\frac{1}{2}}=\frac{1}{3}\left(1+\frac{2}{3}x\right)^{-\frac{1}{2}}$
Expand the bracket using $n=−12n=-\frac{1}{2}$ and $u=23xu=\frac{2}{3}x$ :
$(1+23x)−12=1−x3+x26−5x354+⋯\left(1+\frac{2}{3}x\right)^{-\frac{1}{2}}=1-\frac{x}{3}+\frac{x^2}{6}-\frac{5x^3}{54}+\cdots$
Multiply every term by $13\frac{1}{3}$ :
$(9+6x)−12=13−x9+x218−5x3162+⋯(9+6x)^{-\frac{1}{2}}=\frac{1}{3}-\frac{x}{9}+\frac{x^2}{18}-\frac{5x^3}{162}+\cdots$
The range comes from $∣23x∣<1\left|\frac{2}{3}x\right|<1$ :
$−32<x<32-\frac{3}{2}<x<\frac{3}{2}$

Common Mistake

Expanding the wrong bracket

Do not apply the formula directly to $9 + 6 x$ as though it were $1 + u$ . The constant factor must come out first.

5. Validity and accuracy

For fractional or negative powers, the expansion is valid when

∣ u ∣ < 1

where $u$ is the small expression in $1+u)^n$ .

For example, if you have $(1−2x9)12\left(1-\frac{2x}{9}\right)^{\frac{1}{2}}$ , then $u=−2x9u=-\frac{2x}{9}$ , so the condition is $∣−2x9∣<1\left|-\frac{2x}{9}\right|<1$ .

This number line shows the range of x values for which the expansion of (1−2x/9)^(1/2) is valid.

Common Mistake

Validity before accuracy

If $∣u∣≥1|u|\ge 1$ , the expansion should not be used. Even when it is valid, it is usually more accurate when $u$ is close to 0.

6. Using expansions for approximations

A binomial expansion is especially useful when the value of $x$ is small. Higher powers like $x^2$ and $x^3$ become much smaller, so the first few terms give a good approximation.

For small x, successive powers shrink rapidly, so later terms usually contribute much less to the approximation.

Example

Approximating a square root

Use a binomial expansion to estimate $8.8\sqrt{8.8}$ to 5 decimal places.

Write $8.8\sqrt{8.8}$ as part of $(9−2x)12(9-2x)^{\frac{1}{2}}$ :
$9 - 2 x = 8.8$
Solve for the substitution value:
$x = 0.1$
Expand $(9−2x)12(9-2x)^{\frac{1}{2}}$ :
$(9−2x)12=3(1−2x9)12=3−x3−x254−x3486+⋯\begin{aligned} (9-2x)^{\frac{1}{2}} &=3\left(1-\frac{2x}{9}\right)^{\frac{1}{2}}\\ &=3-\frac{x}{3}-\frac{x^2}{54}-\frac{x^3}{486}+\cdots \end{aligned}$
Substitute $x = 0.1$ :
$3−0.13−0.1254−0.13486=2.966479…3-\frac{0.1}{3}-\frac{0.1^2}{54}-\frac{0.1^3}{486}=2.966479\ldots$
Therefore, $8.8≈2.96648\sqrt{8.8}\approx 2.96648$ to 5 decimal places.

7. Combining expansions with other algebra

Sometimes you expand one part first, then multiply by another factor or match coefficients.

Multiplying by another factor

Example

Multiplying after expanding

Find the expansion of $1+x(1−2x)2\frac{1+x}{(1-2x)^2}$ up to and including $x^3$ .

First expand the denominator part:
$(1−2x)−2=1+4x+12x2+32x3+⋯(1-2x)^{-2}=1+4x+12x^2+32x^3+\cdots$
Multiply by $1 + x$ , keeping terms only up to $x^3$ :

A multiplication grid makes clear which products contribute up to x^3 and which are ignored.

$$
(1+x)(1+4x+12x^2+32x^3)=1+5x+16x^2+44x^3+\cdots
$$

3. Terms like $x⋅32x3x\cdot 32x^3$ are in $x^4$ , so they are not needed.

Tip

Only keep what can matter

When multiplying expansions up to $x^3$ , ignore any product whose powers already add to more than 3.

Matching coefficients

Example

Finding unknown constants

The expansion of $a+bx4+x\frac{a+bx}{\sqrt{4+x}}$ begins $2+x+⋯2+x+\cdots$ . Find $a$ and $b$ .

Begin with the reciprocal square root expansion:
$(4+x)−12=12−x16+⋯(4+x)^{-\frac{1}{2}}=\frac{1}{2}-\frac{x}{16}+\cdots$
Multiply by $a + b x$ and keep the constant and $x$ terms:
$)=a2+(b2−a16)x+⋯(a+bx)\left(\frac{1}{2}-\frac{x}{16}+\cdots\right)=\frac{a}{2}+\left(\frac{b}{2}-\frac{a}{16}\right)x+\cdots$
Compare coefficients with $2+x+⋯2+x+\cdots$ :
$a2=2,b2−a16=1\frac{a}{2}=2,\qquad \frac{b}{2}-\frac{a}{16}=1$
Solve the equations: $a = 4$ , then $b2−14=1\frac{b}{2}-\frac{1}{4}=1$ , so $b=52b=\frac{5}{2}$ .

Exam technique

In the exam

Rewrite roots and fractions as powers before expanding.
Factor the bracket into $A(1+u)^n$ if the constant term is not 1.
For fractional or negative powers, always state the range using $∣ u ∣ < 1$ .
In approximation questions, choose the substitution with the smallest valid value of $∣ u ∣$ .
When multiplying expansions, keep only the terms that can contribute to the requested power.

Self review

Check yourself

Can you expand $(1−3x)−12(1-3x)^{-\frac{1}{2}}$ up to and including $x^2$ , and state its validity range?
Can you rewrite $125+10x\frac{1}{\sqrt{25+10x}}$ in the form $A(1+u)^n$ ?
If you multiply an expansion by $1 + 2 x$ , which terms can contribute to the coefficient of $x^3$ ?

Binomial Expansion

What you'll learn

1. The language of binomial expansion

2. Positive integer powers

3. Fractional and negative powers

4. Factoring first

5. Validity and accuracy

6. Using expansions for approximations

7. Combining expansions with other algebra

Multiplying by another factor

Matching coefficients

Recap questions

You've reached the end

Binomial Expansion

What you'll learn

1. The language of binomial expansion

2. Positive integer powers

3. Fractional and negative powers

4. Factoring first

5. Validity and accuracy

6. Using expansions for approximations

7. Combining expansions with other algebra

Multiplying by another factor

Matching coefficients

Recap questions

You've reached the end

Binomial Expansion

What you'll learn

1. The language of binomial expansion

2. Positive integer powers

3. Fractional and negative powers

4. Factoring first

5. Validity and accuracy

6. Using expansions for approximations

7. Combining expansions with other algebra

Multiplying by another factor

Matching coefficients

Recap questions

You've reached the end

Binomial Expansion Revision Guide

Binomial Expansion

What you'll learn

1. The language of binomial expansion

2. Positive integer powers

3. Fractional and negative powers

4. Factoring first

5. Validity and accuracy

6. Using expansions for approximations

7. Combining expansions with other algebra

Multiplying by another factor

Matching coefficients

Recap questions

You've reached the end

Binomial Expansion Revision Guide