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Chapter 5 · Additional Maths · A5 Binomial expansions

Binomial expansions

Multiplying a bracket by itself again and again is slow, and one slip spoils everything. The binomial theorem writes down every term of $(a + b)^n$ at once, from Pascal's triangle for small powers or from binomial coefficients for any positive whole number $n$ . By the end you can expand a bracket, pick out one term or one coefficient, find unknown constants, and use the first few terms to estimate a number. This chapter is not in the G2 syllabus: it is for the G3 Additional Mathematics course only.

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Chapter
5

Figure 1 · Pascal's triangle, rows 0 to 6: row 4 is highlighted

3 outcomes · 1 page · every step shown ✦

Read the triangle

Row $n$ gives the coefficients of $(a + b)^n$ . Each inner number is the sum of the two above it; the power of $a$ falls while the power of $b$ rises.

Use the theorem

$\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$ is the number in front of $a^{n-r}b^r$ . Put brackets round each term before you raise it to a power, so its sign and its coefficient are powered too.

Pick one term

The term with $b^r$ is the $(r+1)$ th term. Set the power of $x$ equal to what you want, solve for $r$ , and work out only that term.

By the end of this chapter

Write out Pascal's triangle and use a row of it to expand $(a + b)^n$ for small $n$ , including a related expansion deduced from one you already have.
Evaluate $n!$ and $\binom{n}{r}$ by the formula and on a calculator, and know what a calculator cannot do.
Expand a bracket with the binomial theorem when the terms have coefficients, a negative second term or a fractional second term, in ascending or descending powers.
Use the general term to find a particular term, the coefficient of a power of $x$ , the term independent of $x$ or an unknown constant.
Find coefficients in a product or in a compound bracket, solve for an unknown index or constant, and use the first few terms to estimate a number and comment on its accuracy.

✶

Start here

Maths In Real Life

Two situations where a binomial expansion does real work. The pictures come first; the maths drawn over them is generated from the same numbers that the solutions use.

$A cube-shaped plastic recycling bin, open at the top, with thick walls, standing on a void-deck floor; the outside edge is marked (x + 2) cm, the wall thickness 1 cm and the inside edge x cm$

(x + 2)\text{ cm}

1\text{ cm}

x\text{ cm}

Maths In Real Life 1

The recycling bin

A cube-shaped plastic recycling bin stands in the void deck of an HDB block. It is open at the top. Its walls and its floor are all $1$ cm thick, so the outside is a cube of edge $(x + 2)$ cm, while the hollow inside is $x$ cm by $x$ cm and $(x + 1)$ cm deep. Expanding the cube of the outside edge is the first step to finding how much plastic the bin contains.

The inside of the bin is $x$ cm by $x$ cm by $(x + 1)$ cm deep, and the outside is a cube of edge $(x + 2)$ cm. Volumes are in cubic centimetres.

(a)Use the binomial theorem to expand $(x + 2)^3$ , the volume of the outside cube, in descending powers of $x$ .
(b)The hollow inside has volume $x^2(x + 1)$ . Find an expression for the volume of plastic in the bin.
(c)Find the volume of plastic when $x = 30$ .

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Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

(a)

\displaystyle (x + 2)^3\;= \textcolor{#010103}{\binom{3}{0}}\textcolor{#010104}{x^3} + \textcolor{#010103}{\binom{3}{1}}\textcolor{#010104}{x^2}\textcolor{#010105}{(2)} + \textcolor{#010103}{\binom{3}{2}}\textcolor{#010104}{x}\textcolor{#010105}{(2)^2} + \textcolor{#010103}{\binom{3}{3}}\textcolor{#010105}{(2)^3}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{(x + 2)^3}\;= \textcolor{#010104}{x^3} + \textcolor{#010103}{3}\textcolor{#010104}{x^2}\textcolor{#010105}{(2)} + \textcolor{#010103}{3}\textcolor{#010104}{x}\textcolor{#010105}{(2)^2} + \textcolor{#010105}{(2)^3}

Work out each binomial coefficient.

\displaystyle \phantom{(x + 2)^3}\;= \textcolor{#010104}{x^3} + \textcolor{#010103}{3}\textcolor{#010104}{x^2}\textcolor{#010105}{(2)} + \textcolor{#010103}{3}\textcolor{#010104}{x}\textcolor{#010105}{(4)} + \textcolor{#010105}{8}

Work out each power.

\displaystyle \phantom{(x + 2)^3}\;= \textcolor{#010106}{x^3 + 6x^2 + 12x + 8}

Multiply out each term and tidy the signs. This is already in descending powers of

x

(b)

\displaystyle W\;= (x + 2)^3 - x^2(x + 1)

Plastic = outside volume minus hollow volume.

\displaystyle \phantom{W}\;= x^3 + 6x^2 + 12x + 8 - x^3 - x^2

Put in the expansion from part (a); the hollow is

x^2 \times x + x^2 \times 1

\displaystyle \phantom{W}\;= 5x^2 + 12x + 8

The

x^3

terms cancel and the

x^2

terms combine.

(c)

\displaystyle W\;= 5(30)^2 + 12(30) + 8

Put

x = 30

into each term.

\displaystyle \phantom{W}\;= 4500 + 360 + 8

Powers first, then the multiplications.

\displaystyle \phantom{W}\;= 4868

Add up.

Answer

(a)

(x + 2)^3 = x^3 + 6x^2 + 12x + 8

(b)

W = 5x^2 + 12x + 8

(c)

W = 4868

cubic centimetres

$A glass jar of coins on a study desk next to a window of an HDB flat, with the amount banked, the yearly rate and the number of years marked$

\$2500

0.5\%\text{ a year}

8\text{ years}

Maths In Real Life 2

The savings jar

Mei Ling empties her coin jar, counts $2500 and pays it into a savings plan that adds $0.5\%$ every year. She wants to know what it will be worth in a few years. The rate is small, so the first few terms of a binomial expansion already give an excellent estimate.

Each $1 in the plan grows to $(1 + x)^{n}$ dollars after $n$ years, where $x$ is the yearly rate as a decimal. Take $n = 8$ .

(a)Expand $(1 + x)^{ 8 }$ in ascending powers of $x$ as far as the term in $x^2$ .
(b)Use $x = 0.005$ to estimate $(1 + x)^{ 8 }$ , the growth of each dollar, to $4$ decimal places. Then find the estimated value of $2500 after $8$ years.
(c)Compare your estimate with the value of $(1.005)^{ 8 }$ from a calculator, find the error, and comment on how accurate the estimate is.

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Colour keybinomial coefficientsecond term and its powerfinal result

(a)

The power of

x

goes up as

r

goes up, so the

3

lowest-power terms come from

r = 0, 1, 2

. The rest are not needed.

\displaystyle T_{1}\;= \textcolor{#010103}{\binom{8}{0}}

Put

r = 0

into the general term.

\displaystyle \phantom{T_{1}}\;= \textcolor{#010106}{1}

Work out the binomial coefficient and the powers.

\displaystyle T_{2}\;= \textcolor{#010103}{\binom{8}{1}}\textcolor{#010105}{x}

Put

r = 1

into the general term.

\displaystyle \phantom{T_{2}}\;= \textcolor{#010106}{8x}

Work out the binomial coefficient and the powers.

\displaystyle T_{3}\;= \textcolor{#010103}{\binom{8}{2}}\textcolor{#010105}{x^2}

Put

r = 2

into the general term.

\displaystyle \phantom{T_{3}}\;= \textcolor{#010106}{28x^2}

Work out the binomial coefficient and the powers.

Put the

3

terms together:

\textcolor{#010106}{1 + 8x + 28x^2} + \cdots

The dots stand for the higher powers we did not need.

(b)

\displaystyle \left(1.005\right)^{8}\;= \left(1 + 0.005\right)^{8}

Write

1.005

1

plus a small number,

0.005

Keep the first

3

terms only:

0.005

is small, so its higher powers are tiny. The estimate is the sum of those terms.

\displaystyle 1 + \textcolor{#010105}{8(0.005)} + \textcolor{#010105}{28(0.005)^2} = 1 + 0.04 + 0.0007 = \textcolor{#010106}{1.0407}

Powers first, then the multiplications, then add up.

\left(1.005\right)^{8} \approx 1.0407

\displaystyle 2500 \times 1.0407 = \textcolor{#010106}{2601.75}

Multiply the starting amount by the estimated growth factor

1.0407

(c)

A calculator gives

\left(1.005\right)^{8} \approx 1.040707

(to

6

decimal places). Compare it with your estimate

1.0407

, the answer correct to

4

decimal places.

\displaystyle 1.040707 - 1.0407 = \textcolor{#010106}{0.000007}

The difference between the two values: the estimate

1.0407

is below the calculator value.

For the money: the estimate gives about $2601.75 and the calculator value gives about $2601.77.

\displaystyle 56 \times (0.005)^{3} = 56 \times 0.000000125 = \textcolor{#010106}{0.000007}

The first term left out: the binomial coefficient is

56

, times

0.005

to the power

3

The estimate and the calculator value both round to

1.0407

(to

4

decimal places), so the estimate is correct to

4

decimal places. This was expected: the first term left out is only

0.000007

, smaller than half a unit of the last place,

0.00005

, and the later terms are smaller still, because each one has another factor of

0.005

Compare the two rounded values first; the size of the next term explains the result.

Answer

(a)

(1 + x)^8 = 1 + 8x + 28x^2 + \cdots

(b)

\left(1.005\right)^{8} \approx 1.0407

; about $2601.75

1.0407

is about

0.000007

below the calculator value; the money is about $2601.75 by the estimate against $2601.77 by calculator; the first term left out is only

0.000007

; the estimate is correct to

4

decimal places, since it and the calculator value both round to

1.0407

Supports K341-A5-01

Pascal's triangle and small powers

Multiplying out $(a + b)^2$ gives $a^2 + 2ab + b^2$ . Do the same for $(a + b)^5$ and you must multiply five brackets together, which is slow and easy to get wrong. The numbers in front of the terms follow a pattern, and that pattern is called Pascal's triangle.

Start with a single $1$ at the top. Every new row begins and ends with $1$ , and every number in between is the sum of the two numbers above it. Rows are counted from $0$ , so row $n$ goes with the power $n$ : row $2$ is $1, 2, 1$ (look at $a^2 + 2ab + b^2$ ) and row $3$ is $1, 3, 3, 1$ .

Rows

0

6

. Row

3

is highlighted:

1, 3, 3, 1

are the coefficients of

(a + b)^3

The Core Idea

Row $n$ of Pascal's triangle gives the coefficients of $(a + b)^n$ .

In every term the power of $a$ falls from $n$ to $0$ while the power of $b$ rises from $0$ to $n$ . The two powers in a term always add up to $n$ , and there are $n + 1$ terms.

Here is the whole method for $(x + 5)^3$ . Row $3$ is $1, 3, 3, 1$ . Write each term as coefficient $\times$ (first term to a falling power) $\times$ (second term to a rising power), then work out the product.

Case table 1: building $(x + 5)^3$ term by term from row $3$
r	Coefficient	First term, falling power	Second term, rising power	The term
$0$	$1$	$x^3$	$1$	$x^3$
$1$	$3$	$x^2$	$(5)$	$15x^2$
$2$	$3$	$x$	$(5)^2$	$75x$
$3$	$1$	$1$	$(5)^3$	$125$

Example 1.1

Build a row of the triangle

Write down row $5$ of Pascal's triangle.

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Row

0

is just

1

. Every row starts and ends with

1

, and every number in between is the sum of the two numbers above it.

The two numbers above sit just left and right of the gap.

Row

2

comes from row

1

(

1, 1

1

1 + 1 = 2

1

Add each neighbouring pair of the row above.

Row

3

comes from row

2

(

1, 2, 1

1

1 + 2 = 3

2 + 1 = 3

1

Row

4

comes from row

3

(

1, 3, 3, 1

1

1 + 3 = 4

3 + 3 = 6

3 + 1 = 4

1

Row

5

comes from row

4

(

1, 4, 6, 4, 1

1

1 + 4 = 5

4 + 6 = 10

6 + 4 = 10

4 + 1 = 5

1

Answer

row

5

1, 5, 10, 10, 5, 1

Your Turn 1.1

Write down row $6$ of Pascal's triangle.

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Row

0

is just

1

. Every row starts and ends with

1

, and every number in between is the sum of the two numbers above it.

The two numbers above sit just left and right of the gap.

Row

2

comes from row

1

(

1, 1

1

1 + 1 = 2

1

Add each neighbouring pair of the row above.

Row

3

comes from row

2

(

1, 2, 1

1

1 + 2 = 3

2 + 1 = 3

1

Row

4

comes from row

3

(

1, 3, 3, 1

1

1 + 3 = 4

3 + 3 = 6

3 + 1 = 4

1

Row

5

comes from row

4

(

1, 4, 6, 4, 1

1

1 + 4 = 5

4 + 6 = 10

6 + 4 = 10

4 + 1 = 5

1

Row

6

comes from row

5

(

1, 5, 10, 10, 5, 1

1

1 + 5 = 6

5 + 10 = 15

10 + 10 = 20

10 + 5 = 15

5 + 1 = 6

1

Answer

row

6

1, 6, 15, 20, 15, 6, 1

Example 1.2

Expand with a positive number

Use Pascal's triangle to expand $(x + 2)^4$ .

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Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

Row

3

of Pascal's triangle is

1, 3, 3, 1

, so row

4

1

1 + 3 = 4

3 + 3 = 6

3 + 1 = 4

1

: that is

1, 4, 6, 4, 1

Each inner number is the sum of the two above it.

(x + 2)^4

the power of the first term falls from

4

0

while the power of the second term rises from

0

4

; the row gives the coefficients in order.

\displaystyle (x + 2)^4\;= \textcolor{#010104}{x^4} + \textcolor{#010103}{4}\textcolor{#010104}{x^3}\textcolor{#010105}{(2)} + \textcolor{#010103}{6}\textcolor{#010104}{x^2}\textcolor{#010105}{(2)^2} + \textcolor{#010103}{4}\textcolor{#010104}{x}\textcolor{#010105}{(2)^3} + \textcolor{#010105}{(2)^4}

Coefficient, then the first term to a falling power, then the second term to a rising power.

\displaystyle \phantom{(x + 2)^4}\;= \textcolor{#010104}{x^4} + \textcolor{#010103}{4}\textcolor{#010104}{x^3}\textcolor{#010105}{(2)} + \textcolor{#010103}{6}\textcolor{#010104}{x^2}\textcolor{#010105}{(4)} + \textcolor{#010103}{4}\textcolor{#010104}{x}\textcolor{#010105}{(8)} + \textcolor{#010105}{16}

Work out each power.

\displaystyle \phantom{(x + 2)^4}\;= \textcolor{#010106}{x^4 + 8x^3 + 24x^2 + 32x + 16}

Multiply out each term and tidy the signs.

Answer

(x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16

Your Turn 1.2

Use Pascal's triangle to expand $(x + 3)^3$ .

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Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

Row

2

of Pascal's triangle is

1, 2, 1

, so row

3

1

1 + 2 = 3

2 + 1 = 3

1

: that is

1, 3, 3, 1

Each inner number is the sum of the two above it.

(x + 3)^3

the power of the first term falls from

3

0

while the power of the second term rises from

0

3

; the row gives the coefficients in order.

\displaystyle (x + 3)^3\;= \textcolor{#010104}{x^3} + \textcolor{#010103}{3}\textcolor{#010104}{x^2}\textcolor{#010105}{(3)} + \textcolor{#010103}{3}\textcolor{#010104}{x}\textcolor{#010105}{(3)^2} + \textcolor{#010105}{(3)^3}

Coefficient, then the first term to a falling power, then the second term to a rising power.

\displaystyle \phantom{(x + 3)^3}\;= \textcolor{#010104}{x^3} + \textcolor{#010103}{3}\textcolor{#010104}{x^2}\textcolor{#010105}{(3)} + \textcolor{#010103}{3}\textcolor{#010104}{x}\textcolor{#010105}{(9)} + \textcolor{#010105}{27}

Work out each power.

\displaystyle \phantom{(x + 3)^3}\;= \textcolor{#010106}{x^3 + 9x^2 + 27x + 27}

Multiply out each term and tidy the signs.

Answer

(x + 3)^3 = x^3 + 9x^2 + 27x + 27

Example 1.3

A negative second term

Use Pascal's triangle to expand $(x - 2)^5$ . Look at the sign of the second term before you start.

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Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

Row

4

of Pascal's triangle is

1, 4, 6, 4, 1

, so row

5

1

1 + 4 = 5

4 + 6 = 10

6 + 4 = 10

4 + 1 = 5

1

: that is

1, 5, 10, 10, 5, 1

Each inner number is the sum of the two above it.

(x - 2)^5

the power of the first term falls from

5

0

while the power of the second term rises from

0

5

; the row gives the coefficients in order.

\displaystyle (x - 2)^5\;= \textcolor{#010104}{x^5} + \textcolor{#010103}{5}\textcolor{#010104}{x^4}\textcolor{#010105}{(-2)} + \textcolor{#010103}{10}\textcolor{#010104}{x^3}\textcolor{#010105}{(-2)^2} + \textcolor{#010103}{10}\textcolor{#010104}{x^2}\textcolor{#010105}{(-2)^3} + \textcolor{#010103}{5}\textcolor{#010104}{x}\textcolor{#010105}{(-2)^4} + \textcolor{#010105}{(-2)^5}

Coefficient, then the first term to a falling power, then the second term to a rising power.

\displaystyle \phantom{(x - 2)^5}\;= \textcolor{#010104}{x^5} + \textcolor{#010103}{5}\textcolor{#010104}{x^4}\textcolor{#010105}{(-2)} + \textcolor{#010103}{10}\textcolor{#010104}{x^3}\textcolor{#010105}{(4)} + \textcolor{#010103}{10}\textcolor{#010104}{x^2}\textcolor{#010105}{(-8)} + \textcolor{#010103}{5}\textcolor{#010104}{x}\textcolor{#010105}{(16)} - \textcolor{#010105}{32}

Work out each power. Keep the minus sign of the second term inside its bracket.

\displaystyle \phantom{(x - 2)^5}\;= \textcolor{#010106}{x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32}

Multiply out each term and tidy the signs.

Answer

(x - 2)^5 = x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32

Your Turn 1.3

Use Pascal's triangle to expand $(x - 1)^6$ . Look at the sign of the second term before you start.

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Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

Row

5

of Pascal's triangle is

1, 5, 10, 10, 5, 1

, so row

6

1

1 + 5 = 6

5 + 10 = 15

10 + 10 = 20

10 + 5 = 15

5 + 1 = 6

1

: that is

1, 6, 15, 20, 15, 6, 1

Each inner number is the sum of the two above it.

(x - 1)^6

the power of the first term falls from

6

0

while the power of the second term rises from

0

6

; the row gives the coefficients in order.

\displaystyle (x - 1)^6\;= \textcolor{#010104}{x^6} + \textcolor{#010103}{6}\textcolor{#010104}{x^5}\textcolor{#010105}{(-1)} + \textcolor{#010103}{15}\textcolor{#010104}{x^4}\textcolor{#010105}{(-1)^2} + \textcolor{#010103}{20}\textcolor{#010104}{x^3}\textcolor{#010105}{(-1)^3} + \textcolor{#010103}{15}\textcolor{#010104}{x^2}\textcolor{#010105}{(-1)^4} + \textcolor{#010103}{6}\textcolor{#010104}{x}\textcolor{#010105}{(-1)^5} + \textcolor{#010105}{(-1)^6}

Coefficient, then the first term to a falling power, then the second term to a rising power.

\displaystyle \phantom{(x - 1)^6}\;= \textcolor{#010104}{x^6} + \textcolor{#010103}{6}\textcolor{#010104}{x^5}\textcolor{#010105}{(-1)} + \textcolor{#010103}{15}\textcolor{#010104}{x^4}\textcolor{#010105}{(1)} + \textcolor{#010103}{20}\textcolor{#010104}{x^3}\textcolor{#010105}{(-1)} + \textcolor{#010103}{15}\textcolor{#010104}{x^2}\textcolor{#010105}{(1)} + \textcolor{#010103}{6}\textcolor{#010104}{x}\textcolor{#010105}{(-1)} + \textcolor{#010105}{1}

Work out each power. Keep the minus sign of the second term inside its bracket.

\displaystyle \phantom{(x - 1)^6}\;= \textcolor{#010106}{x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1}

Multiply out each term and tidy the signs.

Answer

(x - 1)^6 = x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1

Example 1.4

A coefficient on the first term

Use Pascal's triangle to expand $(2x + 1)^4$ .

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Colour keybinomial coefficientfirst term and its powerfinal result

Row

3

of Pascal's triangle is

1, 3, 3, 1

, so row

4

1

1 + 3 = 4

3 + 3 = 6

3 + 1 = 4

1

: that is

1, 4, 6, 4, 1

Each inner number is the sum of the two above it.

(2x + 1)^4

the power of the first term falls from

4

0

while the power of the second term rises from

0

4

; the row gives the coefficients in order.

\displaystyle (2x + 1)^4\;= \textcolor{#010104}{(2x)^4} + \textcolor{#010103}{4}\textcolor{#010104}{(2x)^3} + \textcolor{#010103}{6}\textcolor{#010104}{(2x)^2} + \textcolor{#010103}{4}\textcolor{#010104}{(2x)} + \textcolor{#010103}{1}

Coefficient, then the first term to a falling power, then the second term to a rising power.

\displaystyle \phantom{(2x + 1)^4}\;= \textcolor{#010104}{16x^4} + \textcolor{#010103}{4}\textcolor{#010104}{(8x^3)} + \textcolor{#010103}{6}\textcolor{#010104}{(4x^2)} + \textcolor{#010103}{4}\textcolor{#010104}{(2x)} + \textcolor{#010103}{1}

Work out each power.

\displaystyle \phantom{(2x + 1)^4}\;= \textcolor{#010106}{16x^4 + 32x^3 + 24x^2 + 8x + 1}

Multiply out each term and tidy the signs.

Answer

(2x + 1)^4 = 16x^4 + 32x^3 + 24x^2 + 8x + 1

Your Turn 1.4

Use Pascal's triangle to expand $(3x + 1)^3$ .

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Colour keybinomial coefficientfirst term and its powerfinal result

Row

2

of Pascal's triangle is

1, 2, 1

, so row

3

1

1 + 2 = 3

2 + 1 = 3

1

: that is

1, 3, 3, 1

Each inner number is the sum of the two above it.

(3x + 1)^3

the power of the first term falls from

3

0

while the power of the second term rises from

0

3

; the row gives the coefficients in order.

\displaystyle (3x + 1)^3\;= \textcolor{#010104}{(3x)^3} + \textcolor{#010103}{3}\textcolor{#010104}{(3x)^2} + \textcolor{#010103}{3}\textcolor{#010104}{(3x)} + \textcolor{#010103}{1}

Coefficient, then the first term to a falling power, then the second term to a rising power.

\displaystyle \phantom{(3x + 1)^3}\;= \textcolor{#010104}{27x^3} + \textcolor{#010103}{3}\textcolor{#010104}{(9x^2)} + \textcolor{#010103}{3}\textcolor{#010104}{(3x)} + \textcolor{#010103}{1}

Work out each power.

\displaystyle \phantom{(3x + 1)^3}\;= \textcolor{#010106}{27x^3 + 27x^2 + 9x + 1}

Multiply out each term and tidy the signs.

Answer

(3x + 1)^3 = 27x^3 + 27x^2 + 9x + 1

Example 1.5

Deduce a related expansion

(i)Use Pascal's triangle to expand $(1 + x)^{ 4 }$ .
(ii)Hence expand $(1 + 3y)^4$ in ascending powers of $y$ .

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Colour keysecond term and its powerfinal result

(i)

Row

3

of Pascal's triangle is

1, 3, 3, 1

, so row

4

1

1 + 3 = 4

3 + 3 = 6

3 + 1 = 4

1

: that is

1, 4, 6, 4, 1

Each inner number is the sum of the two above it.

(1 + x)^4

the power of the first term falls from

4

0

while the power of the second term rises from

0

4

; the row gives the coefficients in order.

\displaystyle (1 + x)^4\;= \textcolor{#010106}{1 + 4x + 6x^2 + 4x^3 + x^4}

Coefficient, then the first term to a falling power, then the second term to a rising power.

(ii)

\displaystyle (1 + x)^4\;= 1 + 4x + 6x^2 + 4x^3 + x^4

The expansion we already have.

The result is true for every value of

x

, so we may replace

x

3y

on both sides. Put brackets round the replacement so its sign and coefficient are powered too.

\displaystyle \left(1 + (3y)\right)^4\;= 1 + \textcolor{#010105}{4(3y)} + \textcolor{#010105}{6(3y)^2} + \textcolor{#010105}{4(3y)^3} + \textcolor{#010105}{(3y)^4}

Every

x

becomes the bracket.

\displaystyle \phantom{\left(1 + (3y)\right)^4}\;= \textcolor{#010106}{1 + 12y + 54y^2 + 108y^3 + 81y^4}

Work out each power and tidy the signs.

Answer

(i)

(1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4

(ii)

(1 + 3y)^4 = 1 + 12y + 54y^2 + 108y^3 + 81y^4

Your Turn 1.5

(i)Use Pascal's triangle to expand $(1 + x)^{ 6 }$ .
(ii)Hence expand $(1 - 2y)^6$ in ascending powers of $y$ .

Show full solutionHide solution

Colour keysecond term and its powerfinal result

(i)

Row

5

of Pascal's triangle is

1, 5, 10, 10, 5, 1

, so row

6

1

1 + 5 = 6

5 + 10 = 15

10 + 10 = 20

10 + 5 = 15

5 + 1 = 6

1

: that is

1, 6, 15, 20, 15, 6, 1

Each inner number is the sum of the two above it.

(1 + x)^6

the power of the first term falls from

6

0

while the power of the second term rises from

0

6

; the row gives the coefficients in order.

\displaystyle (1 + x)^6\;= \textcolor{#010106}{1 + 6x + 15x^2 + 20x^3 + 15x^4 + 6x^5 + x^6}

Coefficient, then the first term to a falling power, then the second term to a rising power.

(ii)

\displaystyle (1 + x)^6\;= 1 + 6x + 15x^2 + 20x^3 + 15x^4 + 6x^5 + x^6

The expansion we already have.

The result is true for every value of

x

, so we may replace

x

-2y

on both sides. Put brackets round the replacement so its sign and coefficient are powered too.

\displaystyle \left(1 + (-2y)\right)^6\;= 1 + \textcolor{#010105}{6(-2y)} + \textcolor{#010105}{15(-2y)^2} + \textcolor{#010105}{20(-2y)^3} + \textcolor{#010105}{15(-2y)^4} + \textcolor{#010105}{6(-2y)^5} + \textcolor{#010105}{(-2y)^6}

Every

x

becomes the bracket.

\displaystyle \phantom{\left(1 + (-2y)\right)^6}\;= \textcolor{#010106}{1 - 12y + 60y^2 - 160y^3 + 240y^4 - 192y^5 + 64y^6}

Work out each power and tidy the signs.

Answer

(i)

(1 + x)^6 = 1 + 6x + 15x^2 + 20x^3 + 15x^4 + 6x^5 + x^6

(ii)

(1 - 2y)^6 = 1 - 12y + 60y^2 - 160y^3 + 240y^4 - 192y^5 + 64y^6

Pascal's triangle is quick up to about row $6$ or $7$ . After that it is easier to find a single coefficient from a formula, which is the next section.

Rows are counted from

0

. The row for

(a + b)^4

is the one with

5

numbers, row

4

(not the fourth row you see). A quick check:

(a + b)^n

always has

n + 1

terms.

Think About It

Every row of Pascal's triangle starts and ends with

1

. Why? Think about how the terms

a^4

and

b^4

arise when you expand

(a + b)^4

Show the full workingHide the working

To get

a^4

you must pick the

a

from all four brackets, and there is exactly one way to do that. The same holds for

b^4

: pick

b

every time.

A coefficient counts the ways to make a term. One way each means the first and the last coefficients of every row are

1

Answer

Because

a^n

and

b^n

can each be made in only one way.

SGSchoolKaki Tip

Check before you simplify: the powers in every term should add up to $n$ , and the coefficients should be the row you read off.

Supports K341-A5-02

Factorials and binomial coefficients

A factorial is a product of whole numbers counting down: $n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$ . So $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ . We also agree that $0! = 1$ ; the first Think About It box in this section explains why.

The number in front of a term is a binomial coefficient, written $\binom{n}{r}$ and read 'n choose r'. It is the number in row $n$ of Pascal's triangle at position $r$ , counting from $r = 0$ on the left.

The Core Idea

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

Cancel $(n-r)!$ to get the quicker form $\dfrac{n(n-1)\cdots(n-r+1)}{r!}$ : $r$ factors on the top and $r!$ on the bottom. The ends are always $\binom{n}{0} = \binom{n}{n} = 1$ .

Two small cases are worth knowing by heart: $\binom{n}{1} = n$ and $\binom{n}{2} = \dfrac{n(n-1)}{2}$ . For $\binom{n}{3}$ you need $\dfrac{n(n-1)(n-2)}{6}$ .

Example 2.1

Evaluate a factorial

Evaluate $6!$ .

Show solutionHide solution

Colour keyfinal result

n!

means

n

times every whole number below it, down to

1

n! = n(n-1)(n-2)\cdots 1

\displaystyle 6!\;= 6 \times 5 \times 4 \times 3 \times 2 \times 1 = \textcolor{#010106}{720}

Multiply from the left: each answer is the product so far.

Answer

6! = 720

Example 2.2

Evaluate a binomial coefficient

Use the formula to evaluate $\binom{8}{3}$ .

Show solutionHide solution

Colour keyfinal result

\displaystyle \binom{8}{3}\;= \dfrac{8!}{3!\,(8-3)!}

Put

n

and

r

into the coefficient formula.

\displaystyle \phantom{\binom{8}{3}}\;= \dfrac{8!}{3!\,5!}

Work out

n - r

first:

8-3 = 5

\displaystyle \phantom{\binom{8}{3}}\;= \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1}

Cancel

5!

from the top and the bottom: only the first

3

factors of

8!

stay.

\displaystyle \phantom{\binom{8}{3}}\;= \dfrac{336}{6} = \textcolor{#010106}{56}

Multiply the top, multiply the bottom, divide.

Answer

\binom{8}{3} = 56

Your Turn 2.2

Use the formula to evaluate $\binom{9}{4}$ .

Show full solutionHide solution

Colour keyfinal result

\displaystyle \binom{9}{4}\;= \dfrac{9!}{4!\,(9-4)!}

Put

n

and

r

into the coefficient formula.

\displaystyle \phantom{\binom{9}{4}}\;= \dfrac{9!}{4!\,5!}

Work out

n - r

first:

9-4 = 5

\displaystyle \phantom{\binom{9}{4}}\;= \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}

Cancel

5!

from the top and the bottom: only the first

4

factors of

9!

stay.

\displaystyle \phantom{\binom{9}{4}}\;= \dfrac{3024}{24} = \textcolor{#010106}{126}

Multiply the top, multiply the bottom, divide.

Answer

\binom{9}{4} = 126

Example 2.3

Cancel a factorial ratio

Without a calculator, simplify $\dfrac{ 9! }{ 6! }$ .

Show solutionHide solution

Colour keyfinal result

\displaystyle \dfrac{9!}{6!}\;= \dfrac{9 \times 8 \times 7 \times 6!}{6!} = 9 \times 8 \times 7 = \textcolor{#010106}{504}

Write

9!

as the product down to

7

, then times

6!

, which cancels.

Answer

\dfrac{9!}{6!} = 504

Example 2.4

A whole row of coefficients

Find $\binom{6}{0}, \binom{6}{1}, \ldots, \binom{6}{6}$ and compare them with row $6$ of Pascal's triangle.

Show solutionHide solution

Use the formula for each

r

from

0

6

\binom{6}{0} = 1

\binom{6}{1} = 6

\binom{6}{2} = 15

\binom{6}{3} = 20

\binom{6}{4} = 15

\binom{6}{5} = 6

\binom{6}{6} = 1

These are exactly the numbers in row

6

of Pascal's triangle:

1, 6, 15, 20, 15, 6, 1

Row

n

of the triangle lists

\binom{n}{0}, \binom{n}{1}, \ldots

in order.

Answer

1, 6, 15, 20, 15, 6, 1

Your Turn 2.4

Find $\binom{7}{0}, \binom{7}{1}, \ldots, \binom{7}{7}$ and compare them with row $7$ of Pascal's triangle.

Show full solutionHide solution

Use the formula for each

r

from

0

7

\binom{7}{0} = 1

\binom{7}{1} = 7

\binom{7}{2} = 21

\binom{7}{3} = 35

\binom{7}{4} = 35

\binom{7}{5} = 21

\binom{7}{6} = 7

\binom{7}{7} = 1

These are exactly the numbers in row

7

of Pascal's triangle:

1, 7, 21, 35, 35, 21, 7, 1

Row

n

of the triangle lists

\binom{n}{0}, \binom{n}{1}, \ldots

in order.

Answer

1, 7, 21, 35, 35, 21, 7, 1

Example 2.5

Use the calculator key

Evaluate $\binom{15}{4}$ on your calculator, then check it by cancelling.

Show solutionHide solution

On a scientific calculator, enter

15

, press the nCr key, then enter

4

and press equals. The display shows

1365

Order matters:

n

first, then nCr, then

r

Check with the short form:

\dfrac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = \dfrac{32760}{24} = 1365

Answer

1365

Example 2.6

Brackets with a factorial

Work out $(7 - 3)!$ and also $7 - 3!$ , then explain the difference between the two answers.

Show solutionHide solution

\displaystyle (7-3)!\;= 4! = 24

Brackets first:

7-3 = 4

, then the factorial.

\displaystyle 7-3!\;= 7 - 3 \times 2 \times 1 = 7 - 6 = 1

Without brackets only the

3

gets the factorial.

Answer

(7-3)! = 24

but

7-3! = 1

: the factorial applies only to the number next to it unless brackets say otherwise

Be careful with the calculator. Type the brackets:

(7-3)!

24

, but

7 - 3!

1

because only the

3

gets the factorial. Most scientific calculators also show an error for

n!

once

n

reaches

70

Think About It

The coefficient of

x^5

(1 + x)^5

1

. If the formula for

\binom{5}{0}

is to give

1

, what must

0!

equal?

Show the full workingHide the working

\displaystyle \binom{5}{0}\;= \dfrac{5!}{0!\,(5-0)!}

The formula with

r = 0

\displaystyle \phantom{\binom{5}{0}}\;= \dfrac{5!}{0!\,5!} = \dfrac{1}{0!}

The

5!

cancels and leaves

1

over

0!

The answer must be

1

because the coefficient is

1

, so

\dfrac{1}{0!} = 1

. That only works if

0! = 1

. This is why we define

0! = 1

Answer

0! = 1

, so that

\binom{n}{0} = 1

works.

Think About It

Your calculator can give

\binom{70}{3}

but shows an error for

70!

. Why does one work and not the other?

Show the full workingHide the working

A scientific calculator stops at numbers just under

10^{100}

. Here

69! \approx 1.71 \times 10^{98}

fits, but

70! \approx 1.20 \times 10^{100}

does not.

The calculator does not work out

\binom{70}{3}

by building

70!

first; it cancels as it goes. By hand:

\dfrac{70 \times 69 \times 68}{3 \times 2 \times 1} = \dfrac{328440}{6} = 54740

So for large

n

use the short form: it keeps the numbers small, and it is also quicker by hand.

Answer

70!

is bigger than the calculator can show, but

\binom{70}{3} = 54740

stays small if you cancel.

Beyond the Syllabus · Good to Know

The coefficients are symmetric: $\binom{n}{r} = \binom{n}{n-r}$ , which is why every row of Pascal's triangle reads the same from both ends. For example $\binom{10}{7} = \binom{10}{3} = 120$ , so you may swap to the smaller $r$ to save effort. Another is Pascal's rule, $\\binom{n}{r} = \\binom{n-1}{r-1} + \\binom{n-1}{r}$ , which is the sum-of-the-two-above rule written with coefficients. These are not required: you only need the triangle's building rule from section 1. This is never examined, so use it only if it helps.

SGSchoolKaki Tip

For $\binom{n}{r}$ with a large $n$ , cancel before you multiply. Put the smaller of $r$ and $n - r$ on the bottom and write only that many factors on top.

Supports K341-A5-01

The binomial theorem

Pascal's triangle gets long, but the idea behind it works for any positive whole number $n$ . The coefficient in front of $a^{n-r}b^r$ is the binomial coefficient $\binom{n}{r}$ from the last section.

The Core Idea

$\displaystyle (a + b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b$ $\displaystyle + \binom{n}{2}a^{n-2}b^2 + \cdots$ $\displaystyle + \binom{n}{r}a^{n-r}b^r + \cdots + b^n$

This is the binomial theorem, for $n$ a positive whole number. It is on the exam formula page, so you do not have to memorise it, but you must be fluent with it.

To use it, put brackets round $a$ and $b$ exactly as they appear, with their coefficients and signs, and then raise each bracket to its power. The bracket does the work for you: $(2x)^3$ is $8x^3$ , and $(-3)^2$ is $9$ .

Case table 2: building $(2x - 1)^4$ term by term with the theorem
r	Coefficient	First term, falling power	Second term, rising power	The term
$0$	$\binom{4}{0} = 1$	$(2x)^4$	$1$	$16x^4$
$1$	$\binom{4}{1} = 4$	$(2x)^3$	$(-1)$	$-32x^3$
$2$	$\binom{4}{2} = 6$	$(2x)^2$	$(-1)^2$	$24x^2$
$3$	$\binom{4}{3} = 4$	$(2x)$	$(-1)^3$	$-8x$
$4$	$\binom{4}{4} = 1$	$1$	$(-1)^4$	$1$

Example 3.1

The simplest case: $a = 1$

Use the binomial theorem to expand $(1 + x)^5$ .

Show solutionHide solution

Colour keybinomial coefficientsecond term and its powerfinal result

\displaystyle (1 + x)^5\;= \textcolor{#010103}{\binom{5}{0}} + \textcolor{#010103}{\binom{5}{1}}\textcolor{#010105}{x} + \textcolor{#010103}{\binom{5}{2}}\textcolor{#010105}{x^2} + \textcolor{#010103}{\binom{5}{3}}\textcolor{#010105}{x^3} + \textcolor{#010103}{\binom{5}{4}}\textcolor{#010105}{x^4} + \textcolor{#010103}{\binom{5}{5}}\textcolor{#010105}{x^5}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{(1 + x)^5}\;= \textcolor{#010106}{1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5}

Work out each binomial coefficient.

Answer

(1 + x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5

Your Turn 3.1

Use the binomial theorem to expand $(1 + x)^7$ .

Show full solutionHide solution

Colour keybinomial coefficientsecond term and its powerfinal result

\displaystyle (1 + x)^7\;= \textcolor{#010103}{\binom{7}{0}} + \textcolor{#010103}{\binom{7}{1}}\textcolor{#010105}{x} + \textcolor{#010103}{\binom{7}{2}}\textcolor{#010105}{x^2} + \textcolor{#010103}{\binom{7}{3}}\textcolor{#010105}{x^3} + \textcolor{#010103}{\binom{7}{4}}\textcolor{#010105}{x^4} + \textcolor{#010103}{\binom{7}{5}}\textcolor{#010105}{x^5} + \textcolor{#010103}{\binom{7}{6}}\textcolor{#010105}{x^6} + \textcolor{#010103}{\binom{7}{7}}\textcolor{#010105}{x^7}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{(1 + x)^7}\;= \textcolor{#010106}{1 + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + x^7}

Work out each binomial coefficient.

Answer

(1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + x^7

Example 3.2

A coefficient on the first term

Use the binomial theorem to expand $(2x + 3)^4$ .

Show solutionHide solution

Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

\displaystyle (2x + 3)^4\;= \textcolor{#010103}{\binom{4}{0}}\textcolor{#010104}{(2x)^4} + \textcolor{#010103}{\binom{4}{1}}\textcolor{#010104}{(2x)^3}\textcolor{#010105}{(3)} + \textcolor{#010103}{\binom{4}{2}}\textcolor{#010104}{(2x)^2}\textcolor{#010105}{(3)^2} + \textcolor{#010103}{\binom{4}{3}}\textcolor{#010104}{(2x)}\textcolor{#010105}{(3)^3} + \textcolor{#010103}{\binom{4}{4}}\textcolor{#010105}{(3)^4}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{(2x + 3)^4}\;= \textcolor{#010104}{(2x)^4} + \textcolor{#010103}{4}\textcolor{#010104}{(2x)^3}\textcolor{#010105}{(3)} + \textcolor{#010103}{6}\textcolor{#010104}{(2x)^2}\textcolor{#010105}{(3)^2} + \textcolor{#010103}{4}\textcolor{#010104}{(2x)}\textcolor{#010105}{(3)^3} + \textcolor{#010105}{(3)^4}

Work out each binomial coefficient.

\displaystyle \phantom{(2x + 3)^4}\;= \textcolor{#010104}{16x^4} + \textcolor{#010103}{4}\textcolor{#010104}{(8x^3)}\textcolor{#010105}{(3)} + \textcolor{#010103}{6}\textcolor{#010104}{(4x^2)}\textcolor{#010105}{(9)} + \textcolor{#010103}{4}\textcolor{#010104}{(2x)}\textcolor{#010105}{(27)} + \textcolor{#010105}{81}

Work out each power.

\displaystyle \phantom{(2x + 3)^4}\;= \textcolor{#010106}{16x^4 + 96x^3 + 216x^2 + 216x + 81}

Multiply out each term and tidy the signs.

Answer

(2x + 3)^4 = 16x^4 + 96x^3 + 216x^2 + 216x + 81

Your Turn 3.2

Use the binomial theorem to expand $(3x + 2)^3$ .

Show full solutionHide solution

Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

\displaystyle (3x + 2)^3\;= \textcolor{#010103}{\binom{3}{0}}\textcolor{#010104}{(3x)^3} + \textcolor{#010103}{\binom{3}{1}}\textcolor{#010104}{(3x)^2}\textcolor{#010105}{(2)} + \textcolor{#010103}{\binom{3}{2}}\textcolor{#010104}{(3x)}\textcolor{#010105}{(2)^2} + \textcolor{#010103}{\binom{3}{3}}\textcolor{#010105}{(2)^3}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{(3x + 2)^3}\;= \textcolor{#010104}{(3x)^3} + \textcolor{#010103}{3}\textcolor{#010104}{(3x)^2}\textcolor{#010105}{(2)} + \textcolor{#010103}{3}\textcolor{#010104}{(3x)}\textcolor{#010105}{(2)^2} + \textcolor{#010105}{(2)^3}

Work out each binomial coefficient.

\displaystyle \phantom{(3x + 2)^3}\;= \textcolor{#010104}{27x^3} + \textcolor{#010103}{3}\textcolor{#010104}{(9x^2)}\textcolor{#010105}{(2)} + \textcolor{#010103}{3}\textcolor{#010104}{(3x)}\textcolor{#010105}{(4)} + \textcolor{#010105}{8}

Work out each power.

\displaystyle \phantom{(3x + 2)^3}\;= \textcolor{#010106}{27x^3 + 54x^2 + 36x + 8}

Multiply out each term and tidy the signs.

Answer

(3x + 2)^3 = 27x^3 + 54x^2 + 36x + 8

Example 3.3

A negative second term

Expand $(3x - 2)^5$ . Put the minus sign inside the bracket for the second term.

Show solutionHide solution

Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

\displaystyle (3x - 2)^5\;= \textcolor{#010103}{\binom{5}{0}}\textcolor{#010104}{(3x)^5} + \textcolor{#010103}{\binom{5}{1}}\textcolor{#010104}{(3x)^4}\textcolor{#010105}{(-2)} + \textcolor{#010103}{\binom{5}{2}}\textcolor{#010104}{(3x)^3}\textcolor{#010105}{(-2)^2} + \textcolor{#010103}{\binom{5}{3}}\textcolor{#010104}{(3x)^2}\textcolor{#010105}{(-2)^3} + \textcolor{#010103}{\binom{5}{4}}\textcolor{#010104}{(3x)}\textcolor{#010105}{(-2)^4} + \textcolor{#010103}{\binom{5}{5}}\textcolor{#010105}{(-2)^5}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{(3x - 2)^5}\;= \textcolor{#010104}{(3x)^5} + \textcolor{#010103}{5}\textcolor{#010104}{(3x)^4}\textcolor{#010105}{(-2)} + \textcolor{#010103}{10}\textcolor{#010104}{(3x)^3}\textcolor{#010105}{(-2)^2} + \textcolor{#010103}{10}\textcolor{#010104}{(3x)^2}\textcolor{#010105}{(-2)^3} + \textcolor{#010103}{5}\textcolor{#010104}{(3x)}\textcolor{#010105}{(-2)^4} + \textcolor{#010105}{(-2)^5}

Work out each binomial coefficient.

\displaystyle \phantom{(3x - 2)^5}\;= \textcolor{#010104}{243x^5} + \textcolor{#010103}{5}\textcolor{#010104}{(81x^4)}\textcolor{#010105}{(-2)} + \textcolor{#010103}{10}\textcolor{#010104}{(27x^3)}\textcolor{#010105}{(4)} + \textcolor{#010103}{10}\textcolor{#010104}{(9x^2)}\textcolor{#010105}{(-8)} + \textcolor{#010103}{5}\textcolor{#010104}{(3x)}\textcolor{#010105}{(16)} - \textcolor{#010105}{32}

Work out each power. Keep the minus sign of the second term inside its bracket.

\displaystyle \phantom{(3x - 2)^5}\;= \textcolor{#010106}{243x^5 - 810x^4 + 1080x^3 - 720x^2 + 240x - 32}

Multiply out each term and tidy the signs.

Answer

(3x - 2)^5 = 243x^5 - 810x^4 + 1080x^3 - 720x^2 + 240x - 32

Your Turn 3.3

Expand $(2x - 1)^6$ . Put the minus sign inside the bracket for the second term.

Show full solutionHide solution

Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

\displaystyle (2x - 1)^6\;= \textcolor{#010103}{\binom{6}{0}}\textcolor{#010104}{(2x)^6} + \textcolor{#010103}{\binom{6}{1}}\textcolor{#010104}{(2x)^5}\textcolor{#010105}{(-1)} + \textcolor{#010103}{\binom{6}{2}}\textcolor{#010104}{(2x)^4}\textcolor{#010105}{(-1)^2} + \textcolor{#010103}{\binom{6}{3}}\textcolor{#010104}{(2x)^3}\textcolor{#010105}{(-1)^3} + \textcolor{#010103}{\binom{6}{4}}\textcolor{#010104}{(2x)^2}\textcolor{#010105}{(-1)^4} + \textcolor{#010103}{\binom{6}{5}}\textcolor{#010104}{(2x)}\textcolor{#010105}{(-1)^5} + \textcolor{#010103}{\binom{6}{6}}\textcolor{#010105}{(-1)^6}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{(2x - 1)^6}\;= \textcolor{#010104}{(2x)^6} + \textcolor{#010103}{6}\textcolor{#010104}{(2x)^5}\textcolor{#010105}{(-1)} + \textcolor{#010103}{15}\textcolor{#010104}{(2x)^4}\textcolor{#010105}{(-1)^2} + \textcolor{#010103}{20}\textcolor{#010104}{(2x)^3}\textcolor{#010105}{(-1)^3} + \textcolor{#010103}{15}\textcolor{#010104}{(2x)^2}\textcolor{#010105}{(-1)^4} + \textcolor{#010103}{6}\textcolor{#010104}{(2x)}\textcolor{#010105}{(-1)^5} + \textcolor{#010105}{(-1)^6}

Work out each binomial coefficient.

\displaystyle \phantom{(2x - 1)^6}\;= \textcolor{#010104}{64x^6} + \textcolor{#010103}{6}\textcolor{#010104}{(32x^5)}\textcolor{#010105}{(-1)} + \textcolor{#010103}{15}\textcolor{#010104}{(16x^4)}\textcolor{#010105}{(1)} + \textcolor{#010103}{20}\textcolor{#010104}{(8x^3)}\textcolor{#010105}{(-1)} + \textcolor{#010103}{15}\textcolor{#010104}{(4x^2)}\textcolor{#010105}{(1)} + \textcolor{#010103}{6}\textcolor{#010104}{(2x)}\textcolor{#010105}{(-1)} + \textcolor{#010105}{1}

Work out each power. Keep the minus sign of the second term inside its bracket.

\displaystyle \phantom{(2x - 1)^6}\;= \textcolor{#010106}{64x^6 - 192x^5 + 240x^4 - 160x^3 + 60x^2 - 12x + 1}

Multiply out each term and tidy the signs.

Answer

(2x - 1)^6 = 64x^6 - 192x^5 + 240x^4 - 160x^3 + 60x^2 - 12x + 1

Example 3.4

A fractional second term

Expand $\left(x^2 - \dfrac{1}{x}\right)^4$ and simplify each term.

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Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

\displaystyle \left(x^2 - \dfrac{1}{x}\right)^4\;= \textcolor{#010103}{\binom{4}{0}}\textcolor{#010104}{(x^2)^4} + \textcolor{#010103}{\binom{4}{1}}\textcolor{#010104}{(x^2)^3}\textcolor{#010105}{\left(-\dfrac{1}{x}\right)} + \textcolor{#010103}{\binom{4}{2}}\textcolor{#010104}{(x^2)^2}\textcolor{#010105}{\left(-\dfrac{1}{x}\right)^2} + \textcolor{#010103}{\binom{4}{3}}\textcolor{#010104}{(x^2)}\textcolor{#010105}{\left(-\dfrac{1}{x}\right)^3} + \textcolor{#010103}{\binom{4}{4}}\textcolor{#010105}{\left(-\dfrac{1}{x}\right)^4}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{\left(x^2 - \dfrac{1}{x}\right)^4}\;= \textcolor{#010104}{(x^2)^4} + \textcolor{#010103}{4}\textcolor{#010104}{(x^2)^3}\textcolor{#010105}{\left(-\dfrac{1}{x}\right)} + \textcolor{#010103}{6}\textcolor{#010104}{(x^2)^2}\textcolor{#010105}{\left(-\dfrac{1}{x}\right)^2} + \textcolor{#010103}{4}\textcolor{#010104}{(x^2)}\textcolor{#010105}{\left(-\dfrac{1}{x}\right)^3} + \textcolor{#010105}{\left(-\dfrac{1}{x}\right)^4}

Work out each binomial coefficient.

\displaystyle \phantom{\left(x^2 - \dfrac{1}{x}\right)^4}\;= \textcolor{#010104}{x^8} + \textcolor{#010103}{4}\textcolor{#010104}{x^6}\textcolor{#010105}{\left(-\dfrac{1}{x}\right)} + \textcolor{#010103}{6}\textcolor{#010104}{x^4}\textcolor{#010105}{\left(\dfrac{1}{x^2}\right)} + \textcolor{#010103}{4}\textcolor{#010104}{x^2}\textcolor{#010105}{\left(-\dfrac{1}{x^3}\right)} + \textcolor{#010105}{\dfrac{1}{x^4}}

Work out each power. Keep the minus sign of the second term inside its bracket.

\displaystyle \phantom{\left(x^2 - \dfrac{1}{x}\right)^4}\;= \textcolor{#010106}{x^8 - 4x^5 + 6x^2 - \dfrac{4}{x} + \dfrac{1}{x^4}}

Multiply out each term and tidy the signs.

Answer

\left(x^2 - \dfrac{1}{x}\right)^4 = x^8 - 4x^5 + 6x^2 - \dfrac{4}{x} + \dfrac{1}{x^4}

Your Turn 3.4

Expand $\left(x - \dfrac{2}{x^2}\right)^5$ and simplify each term.

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Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

\displaystyle \left(x - \dfrac{2}{x^2}\right)^5\;= \textcolor{#010103}{\binom{5}{0}}\textcolor{#010104}{x^5} + \textcolor{#010103}{\binom{5}{1}}\textcolor{#010104}{x^4}\textcolor{#010105}{\left(-\dfrac{2}{x^2}\right)} + \textcolor{#010103}{\binom{5}{2}}\textcolor{#010104}{x^3}\textcolor{#010105}{\left(-\dfrac{2}{x^2}\right)^2} + \textcolor{#010103}{\binom{5}{3}}\textcolor{#010104}{x^2}\textcolor{#010105}{\left(-\dfrac{2}{x^2}\right)^3} + \textcolor{#010103}{\binom{5}{4}}\textcolor{#010104}{x}\textcolor{#010105}{\left(-\dfrac{2}{x^2}\right)^4} + \textcolor{#010103}{\binom{5}{5}}\textcolor{#010105}{\left(-\dfrac{2}{x^2}\right)^5}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{\left(x - \dfrac{2}{x^2}\right)^5}\;= \textcolor{#010104}{x^5} + \textcolor{#010103}{5}\textcolor{#010104}{x^4}\textcolor{#010105}{\left(-\dfrac{2}{x^2}\right)} + \textcolor{#010103}{10}\textcolor{#010104}{x^3}\textcolor{#010105}{\left(-\dfrac{2}{x^2}\right)^2} + \textcolor{#010103}{10}\textcolor{#010104}{x^2}\textcolor{#010105}{\left(-\dfrac{2}{x^2}\right)^3} + \textcolor{#010103}{5}\textcolor{#010104}{x}\textcolor{#010105}{\left(-\dfrac{2}{x^2}\right)^4} + \textcolor{#010105}{\left(-\dfrac{2}{x^2}\right)^5}

Work out each binomial coefficient.

\displaystyle \phantom{\left(x - \dfrac{2}{x^2}\right)^5}\;= \textcolor{#010104}{x^5} + \textcolor{#010103}{5}\textcolor{#010104}{x^4}\textcolor{#010105}{\left(-\dfrac{2}{x^2}\right)} + \textcolor{#010103}{10}\textcolor{#010104}{x^3}\textcolor{#010105}{\left(\dfrac{4}{x^4}\right)} + \textcolor{#010103}{10}\textcolor{#010104}{x^2}\textcolor{#010105}{\left(-\dfrac{8}{x^6}\right)} + \textcolor{#010103}{5}\textcolor{#010104}{x}\textcolor{#010105}{\left(\dfrac{16}{x^8}\right)} - \textcolor{#010105}{\dfrac{32}{x^{10}}}

Work out each power. Keep the minus sign of the second term inside its bracket.

\displaystyle \phantom{\left(x - \dfrac{2}{x^2}\right)^5}\;= \textcolor{#010106}{x^5 - 10x^2 + \dfrac{40}{x} - \dfrac{80}{x^4} + \dfrac{80}{x^7} - \dfrac{32}{x^{10}}}

Multiply out each term and tidy the signs.

Answer

\left(x - \dfrac{2}{x^2}\right)^5 = x^5 - 10x^2 + \dfrac{40}{x} - \dfrac{80}{x^4} + \dfrac{80}{x^7} - \dfrac{32}{x^{10}}

Example 3.5

Ascending powers

Expand $(3 - 2x)^4$ in ascending powers of $x$ .

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Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

\displaystyle (3 - 2x)^4\;= \textcolor{#010103}{\binom{4}{0}}\textcolor{#010104}{(3)^4} + \textcolor{#010103}{\binom{4}{1}}\textcolor{#010104}{(3)^3}\textcolor{#010105}{(-2x)} + \textcolor{#010103}{\binom{4}{2}}\textcolor{#010104}{(3)^2}\textcolor{#010105}{(-2x)^2} + \textcolor{#010103}{\binom{4}{3}}\textcolor{#010104}{(3)}\textcolor{#010105}{(-2x)^3} + \textcolor{#010103}{\binom{4}{4}}\textcolor{#010105}{(-2x)^4}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{(3 - 2x)^4}\;= \textcolor{#010104}{(3)^4} + \textcolor{#010103}{4}\textcolor{#010104}{(3)^3}\textcolor{#010105}{(-2x)} + \textcolor{#010103}{6}\textcolor{#010104}{(3)^2}\textcolor{#010105}{(-2x)^2} + \textcolor{#010103}{4}\textcolor{#010104}{(3)}\textcolor{#010105}{(-2x)^3} + \textcolor{#010105}{(-2x)^4}

Work out each binomial coefficient.

\displaystyle \phantom{(3 - 2x)^4}\;= \textcolor{#010104}{81} + \textcolor{#010103}{4}\textcolor{#010104}{(27)}\textcolor{#010105}{(-2x)} + \textcolor{#010103}{6}\textcolor{#010104}{(9)}\textcolor{#010105}{(4x^2)} + \textcolor{#010103}{4}\textcolor{#010104}{(3)}\textcolor{#010105}{(-8x^3)} + \textcolor{#010105}{16x^4}

Work out each power. Keep the minus sign of the second term inside its bracket.

\displaystyle \phantom{(3 - 2x)^4}\;= \textcolor{#010106}{81 - 216x + 216x^2 - 96x^3 + 16x^4}

Multiply out each term and tidy the signs. This is already in ascending powers of

x

Answer

(3 - 2x)^4 = 81 - 216x + 216x^2 - 96x^3 + 16x^4

Your Turn 3.5

Expand $(2 + 3x)^5$ in ascending powers of $x$ .

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Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

\displaystyle (2 + 3x)^5\;= \textcolor{#010103}{\binom{5}{0}}\textcolor{#010104}{(2)^5} + \textcolor{#010103}{\binom{5}{1}}\textcolor{#010104}{(2)^4}\textcolor{#010105}{(3x)} + \textcolor{#010103}{\binom{5}{2}}\textcolor{#010104}{(2)^3}\textcolor{#010105}{(3x)^2} + \textcolor{#010103}{\binom{5}{3}}\textcolor{#010104}{(2)^2}\textcolor{#010105}{(3x)^3} + \textcolor{#010103}{\binom{5}{4}}\textcolor{#010104}{(2)}\textcolor{#010105}{(3x)^4} + \textcolor{#010103}{\binom{5}{5}}\textcolor{#010105}{(3x)^5}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{(2 + 3x)^5}\;= \textcolor{#010104}{(2)^5} + \textcolor{#010103}{5}\textcolor{#010104}{(2)^4}\textcolor{#010105}{(3x)} + \textcolor{#010103}{10}\textcolor{#010104}{(2)^3}\textcolor{#010105}{(3x)^2} + \textcolor{#010103}{10}\textcolor{#010104}{(2)^2}\textcolor{#010105}{(3x)^3} + \textcolor{#010103}{5}\textcolor{#010104}{(2)}\textcolor{#010105}{(3x)^4} + \textcolor{#010105}{(3x)^5}

Work out each binomial coefficient.

\displaystyle \phantom{(2 + 3x)^5}\;= \textcolor{#010104}{32} + \textcolor{#010103}{5}\textcolor{#010104}{(16)}\textcolor{#010105}{(3x)} + \textcolor{#010103}{10}\textcolor{#010104}{(8)}\textcolor{#010105}{(9x^2)} + \textcolor{#010103}{10}\textcolor{#010104}{(4)}\textcolor{#010105}{(27x^3)} + \textcolor{#010103}{5}\textcolor{#010104}{(2)}\textcolor{#010105}{(81x^4)} + \textcolor{#010105}{243x^5}

Work out each power.

\displaystyle \phantom{(2 + 3x)^5}\;= \textcolor{#010106}{32 + 240x + 720x^2 + 1080x^3 + 810x^4 + 243x^5}

Multiply out each term and tidy the signs. This is already in ascending powers of

x

Answer

(2 + 3x)^5 = 32 + 240x + 720x^2 + 1080x^3 + 810x^4 + 243x^5

Example 3.6

Descending powers from a constant first term

Expand $(2 + x)^5$ in descending powers of $x$ .

Show solutionHide solution

Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

\displaystyle (2 + x)^5\;= \textcolor{#010103}{\binom{5}{0}}\textcolor{#010104}{(2)^5} + \textcolor{#010103}{\binom{5}{1}}\textcolor{#010104}{(2)^4}\textcolor{#010105}{x} + \textcolor{#010103}{\binom{5}{2}}\textcolor{#010104}{(2)^3}\textcolor{#010105}{x^2} + \textcolor{#010103}{\binom{5}{3}}\textcolor{#010104}{(2)^2}\textcolor{#010105}{x^3} + \textcolor{#010103}{\binom{5}{4}}\textcolor{#010104}{(2)}\textcolor{#010105}{x^4} + \textcolor{#010103}{\binom{5}{5}}\textcolor{#010105}{x^5}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{(2 + x)^5}\;= \textcolor{#010104}{(2)^5} + \textcolor{#010103}{5}\textcolor{#010104}{(2)^4}\textcolor{#010105}{x} + \textcolor{#010103}{10}\textcolor{#010104}{(2)^3}\textcolor{#010105}{x^2} + \textcolor{#010103}{10}\textcolor{#010104}{(2)^2}\textcolor{#010105}{x^3} + \textcolor{#010103}{5}\textcolor{#010104}{(2)}\textcolor{#010105}{x^4} + \textcolor{#010105}{x^5}

Work out each binomial coefficient.

\displaystyle \phantom{(2 + x)^5}\;= \textcolor{#010104}{32} + \textcolor{#010103}{5}\textcolor{#010104}{(16)}\textcolor{#010105}{x} + \textcolor{#010103}{10}\textcolor{#010104}{(8)}\textcolor{#010105}{x^2} + \textcolor{#010103}{10}\textcolor{#010104}{(4)}\textcolor{#010105}{x^3} + \textcolor{#010103}{5}\textcolor{#010104}{(2)}\textcolor{#010105}{x^4} + \textcolor{#010105}{x^5}

Work out each power.

\displaystyle \phantom{(2 + x)^5}\;= \textcolor{#010106}{32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5}

Multiply out each term and tidy the signs.

\displaystyle \phantom{(2 + x)^5}\;= \textcolor{#010106}{x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32}

The question asks for descending powers of

x

, so write the terms in the opposite order.

Answer

(2 + x)^5 = x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32

Your Turn 3.6

Expand $(3 - x)^6$ in descending powers of $x$ .

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Colour keybinomial coefficientfirst term and its powersecond term and its powerfinal result

\displaystyle (3 - x)^6\;= \textcolor{#010103}{\binom{6}{0}}\textcolor{#010104}{(3)^6} + \textcolor{#010103}{\binom{6}{1}}\textcolor{#010104}{(3)^5}\textcolor{#010105}{(-x)} + \textcolor{#010103}{\binom{6}{2}}\textcolor{#010104}{(3)^4}\textcolor{#010105}{(-x)^2} + \textcolor{#010103}{\binom{6}{3}}\textcolor{#010104}{(3)^3}\textcolor{#010105}{(-x)^3} + \textcolor{#010103}{\binom{6}{4}}\textcolor{#010104}{(3)^2}\textcolor{#010105}{(-x)^4} + \textcolor{#010103}{\binom{6}{5}}\textcolor{#010104}{(3)}\textcolor{#010105}{(-x)^5} + \textcolor{#010103}{\binom{6}{6}}\textcolor{#010105}{(-x)^6}

\binom{n}{r}

for every

r

from

0

n

: the first term's power falls, the second term's power rises.

\displaystyle \phantom{(3 - x)^6}\;= \textcolor{#010104}{(3)^6} + \textcolor{#010103}{6}\textcolor{#010104}{(3)^5}\textcolor{#010105}{(-x)} + \textcolor{#010103}{15}\textcolor{#010104}{(3)^4}\textcolor{#010105}{(-x)^2} + \textcolor{#010103}{20}\textcolor{#010104}{(3)^3}\textcolor{#010105}{(-x)^3} + \textcolor{#010103}{15}\textcolor{#010104}{(3)^2}\textcolor{#010105}{(-x)^4} + \textcolor{#010103}{6}\textcolor{#010104}{(3)}\textcolor{#010105}{(-x)^5} + \textcolor{#010105}{(-x)^6}

Work out each binomial coefficient.

\displaystyle \phantom{(3 - x)^6}\;= \textcolor{#010104}{729} + \textcolor{#010103}{6}\textcolor{#010104}{(243)}\textcolor{#010105}{(-x)} + \textcolor{#010103}{15}\textcolor{#010104}{(81)}\textcolor{#010105}{x^2} + \textcolor{#010103}{20}\textcolor{#010104}{(27)}\textcolor{#010105}{(-x^3)} + \textcolor{#010103}{15}\textcolor{#010104}{(9)}\textcolor{#010105}{x^4} + \textcolor{#010103}{6}\textcolor{#010104}{(3)}\textcolor{#010105}{(-x^5)} + \textcolor{#010105}{x^6}

Work out each power. Keep the minus sign of the second term inside its bracket.

\displaystyle \phantom{(3 - x)^6}\;= \textcolor{#010106}{729 - 1458x + 1215x^2 - 540x^3 + 135x^4 - 18x^5 + x^6}

Multiply out each term and tidy the signs.

\displaystyle \phantom{(3 - x)^6}\;= \textcolor{#010106}{x^6 - 18x^5 + 135x^4 - 540x^3 + 1215x^2 - 1458x + 729}

The question asks for descending powers of

x

, so write the terms in the opposite order.

Answer

(3 - x)^6 = x^6 - 18x^5 + 135x^4 - 540x^3 + 1215x^2 - 1458x + 729

The commonest slip is to forget that the power applies to the whole term.

(2x)^3 = 8x^3

, not

2x^3

, and

(-3)^2 = 9

, not

-9

. Write the brackets first and work them out second.

Think About It

When the second term is negative, the signs in the expansion alternate: plus, minus, plus, minus. Show why, using

(x - 2)^3

as the example.

Show the full workingHide the working

The second term is

-2

, and it is raised to the power

r

in the

(r+1)

th term. So the sign of that term is the sign of

(-2)^r

(-2)^0 = 1

(-2)^1 = -2

(-2)^2 = 4

and

(-2)^3 = -8

: even powers are positive and odd powers are negative, so the signs of the terms alternate.

This is why

(x - 2)^3 = x^3 - 6x^2 + 12x - 8

has signs

+, -, +, -

Answer

Because

(-b)^r

is positive for even

r

and negative for odd

r

Links To

A binomial expansion is a polynomial, so everything in Chapter 4 about dividing it and finding its factors applies to it. See Chapter 4: Polynomials, cubic identities and partial fractions

SGSchoolKaki Tip

If the question says 'ascending powers of $x$ ', the lowest power of $x$ comes first. When the first bracket term has a power of $x$ the natural order is descending, so you may need to write the terms in reverse.

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Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 5 of 17