The Binomial Expansion

Expand a bracket to a positive whole-number power with the binomial theorem, find a chosen term or coefficient, and find unknowns from coefficients.

In the exam for IGCSE Additional Maths 0606, Edexcel Further Pure Maths 4PM1 · 14 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

A pattern in the brackets

Multiplying out $(x + 2)^{5}$ means five brackets and a lot of work. But the numbers that appear follow a beautiful pattern. Once you see it, you can expand any power quickly.

Expand $(1 + x)^{n}$ for small $n$ and write down just the numbers in front. They make a triangle:

$n = 1$ : $11$
$n = 2$ : $121$
$n = 3$ : $1331$
$n = 4$ : $14641$
$n = 5$ : $15101051$

Rule

Each row starts and ends with 1. Each number inside a row is the sum of the two numbers just above it. This is Pascal's triangle.

Worked example

1
Expand $(1 + x)^{4}$ .
2
$14641$ — Take the row for $n = 4$ .
3
$1 + 4 x + 6 x^{2} + 4 x^{3} + x^{4}$ — The power of $x$ goes up by one each term, from 0 to 4.

Key idea

In China this pattern is called Yang Hui's triangle, after a mathematician who wrote about it in 1261. That is almost 400 years before Blaise Pascal studied it in France.

Key idea

Row $n$ of the triangle gives the numbers for power $n$ . The powers of $x$ count up from 0 to $n$ .

The binomial theorem

For big powers, drawing the triangle takes too long. Each number in it can be worked out directly.

Formula

(r n) = \frac{n !}{r ! ( n - r )!}

$n$ — the power of the bracket: a positive whole number.
$r$ — which term you want, counting from $r = 0$ for the first term.
$n!$ — n factorial: $n \times (n - 1) \times \dots \times 2 \times 1$ .

Example

No calculator (0606 Paper 1)? Cancel first: $(3 8) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ . On a calculator, use the $^{n} C_{r}$ key.

Formula

(a + b)^{n} = a^{n} + (1 n) a^{n - 1} b + (2 n) a^{n - 2} b^{2} + \dots + b^{n}

$a$ — the first term in the bracket. Its power goes down by 1 each time.
$b$ — the second term, with its sign. Its power goes up by 1 each time.

Rule

This is given on both formula sheets: 0606 and 4037 print it in the List of formulas, and 4PM1 prints the $(1 + x)^{n}$ form you meet in the next step.

Worked example

1
Find the first three terms of $(2 - 3 x)^{5}$ , in ascending powers of $x$ .
2
$a = 2, b = - 3 x, n = 5$ — Name the parts. The minus sign belongs to b.
3
$2^{5} = 32$ — First term.
4
$(1 5) (2)^{4} (- 3 x) = 5 \times 16 \times (- 3 x)$ $= - 240 x$ — Second term: the power of 2 drops to 4, and $(- 3 x)$ appears once.
5
$(2 5) (2)^{3} (- 3 x)^{2} = 10 \times 8 \times 9 x^{2}$ $= 720 x^{2}$ — Third term: $(- 3 x)^{2} = 9 x^{2}$ . Square the 3 and the minus sign.
6
$32 - 240 x + 720 x^{2}$ — “Ascending powers” means start from the number, then $x$ , then $x^{2}$ .

Common mistake

Writing $- 3 x^{2}$ for $(- 3 x)^{2}$ . Keep b in brackets: the number, the sign and the $x$ are all squared.

Key idea

The powers of a and b always add up to $n$ . Use this to check every term.

The (1 + x)ⁿ form

When the bracket starts with 1, all the powers of $a$ are just 1. The theorem becomes a neat series, written without the $(r n)$ symbol.

Formula

(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \frac{n ( n - 1 ) ( n - 2 )}{3 !} x^{3} + \dots

$n$ — the power. For a positive whole number $n$ , the series stops after the $x^{n}$ term.
$x$ — the second term of the bracket. If it is $2 x$ or $a x$ , put the whole thing in brackets.

It is the same theorem: $(2 n) = \frac{n ( n - 1 )}{2 !}$ . This form is useful when $n$ or the number in front of $x$ is unknown.

Worked example

1
Expand $(1 + 2 x)^{7}$ up to the term in $x^{3}$ .
2
$1 + 7 (2 x)$ — First two terms: $n = 7$ , and $x$ is replaced by $2 x$ .
3
$\frac{7 \times 6}{2} (2 x)^{2} = 21 \times 4 x^{2} = 84 x^{2}$ — Third term. $(2 x)^{2} = 4 x^{2}$ .
4
$\frac{7 \times 6 \times 5}{6} (2 x)^{3} = 35 \times 8 x^{3}$ $= 280 x^{3}$ — Fourth term. $3! = 6$ .
5
$1 + 14 x + 84 x^{2} + 280 x^{3}$ — Put the terms together.

Worked example

1
In the expansion of $(1 + a x)^{n}$ , the coefficient of $x$ is 12 and the coefficient of $x^{2}$ is 60. Find $a$ and $n$ .
2
$na = 12$ — The $x$ term is $n (a x)$ .
3
$\frac{n ( n - 1 )}{2} a^{2} = 60$ — The $x^{2}$ term has $(a x)^{2} = a^{2} x^{2}$ .
4
$\frac{( n - 1 ) a \times na}{2} = 60$ $\Rightarrow (n - 1) a \times 6 = 60$ — Split $n (n - 1) a^{2}$ as $(n - 1) a \times na$ , then use $na = 12$ .
5
$na - a = 10 \Rightarrow 12 - a = 10$ — Divide by 6: $(n - 1) a = 10$ . Expand the bracket, then use $na = 12$ .
6
$a = 2, n = 6$ — Then $n = 12 \div 2$ .

Common mistake

Forgetting to square $a$ . The $x^{2}$ term has $(a x)^{2} = a^{2} x^{2}$ , not $a x^{2}$ .

Key idea

Coefficient questions: write each term with the unknowns, set it equal to the number given, then solve the equations together.

One term only: the general term

Rule

This step is for 0606 and 4037.

Sometimes you need just one term, such as the term in $x^{5}$ or the term with no $x$ at all. Write the general term, then choose $r$ .

Formula

T = (r n) a^{n - r} b^{r}

$r$ — a whole number from 0 to $n$ . It is the power of b.
$a$ — the first term in the bracket.
$b$ — the second term in the bracket.

Worked example

1
Find the term independent of $x$ in the expansion of $(x^{2} + \frac{2}{x})^{6}$ .
2
$(r 6) (x^{2})^{6 - r} (\frac{2}{x})^{r}$ — Write the general term. “Independent of $x$ ” means the term with no $x$ .
3
$x^{12 - 2 r} \times x^{- r} = x^{12 - 3 r}$ — Collect the powers of $x$ : $(x^{2})^{6 - r}$ gives $12 - 2 r$ , and $\frac{1}{x ^{r}}$ gives $- r$ .
4
$12 - 3 r = 0, r = 4$ — No $x$ means the power is 0.
5
$(4 6) \times 2^{4} = 15 \times 16 = 240$ — Put $r = 4$ back in. The term is 240.

Example

“Descending powers of $x$ ” means start with the highest power. For $(x + \frac{2}{x})^{4}$ the first three terms are $x^{4} + 8 x^{2} + 24$ .

Mark schemes give a mark for the general term, a mark for the equation in $r$ , then the answer. So write all three lines.

Common mistake

Forgetting that a power of a power multiplies. $(x^{2})^{6 - r}$ is $x^{12 - 2 r}$ , not $x^{8 - r}$ .

Key idea

General term, collect the powers of $x$ , solve for $r$ , then put $r$ back.

Two brackets multiplied

Questions often ask for one coefficient in a product such as $(1 + 3 x) (2 - x)^{5}$ . You do not need the whole expansion. Find every way to make the power you want.

Steps

1
1. Expand the binomial bracket only as far as the power you need.
2. List each pair of terms, one from each bracket, whose powers add up to the power you want.
3. Multiply each pair, then add the results.

Worked example

1
Find the coefficient of $x^{2}$ in $(1 + 3 x) (2 - x)^{5}$ .
2
$(2 - x)^{5} = 32 - 80 x + 80 x^{2} + \dots$ — Only the terms up to $x^{2}$ are needed.
3
$1 \times 80 x^{2} = 80 x^{2}$ — Pair 1: the 1 from the first bracket with the $x^{2}$ term.
4
$3 x \times (- 80 x) = - 240 x^{2}$ — Pair 2: the $3 x$ with the $x$ term.
5
$80 - 240 = - 160$ — Add. The coefficient of $x^{2}$ is −160.

Common mistake

Using only one pair. $1 \times 80 x^{2}$ alone misses the $3 x \times (- 80 x)$ pair.

Key idea

Count the ways to reach the power you want, and add every one.

Which method? Read the question

The wording tells you how much to expand and how.

You see	You do
“Expand” a small power	Pascal's triangle row, or the theorem
“First three terms”, ascending powers	Start from the number term, then $x$ , then $x^{2}$
Descending powers (0606 / 4037)	Start from the highest power of $x$
$(1 + a x)^{n}$ with unknowns	Write the terms with $a$ and $n$ ; set coefficients equal; solve
“Term independent of $x$ ” (0606 / 4037)	General term; power of $x$ = 0; find $r$
A coefficient in a product of brackets	List the pairs that make that power; add

Common mistake

Leaving coefficients unsimplified. Questions ask for each term in its simplest form.

Key idea

Write b with its sign in brackets, check the powers add to $n$ , and simplify every coefficient.

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