1.6 - Expanding integer powers | Cambridge A-Level Mathematics

Multiplying a bracket by itself many times produces a pattern you can use without carrying out every multiplication. In this part of 1.6, you will build the binomial coefficients, expand positive integer powers and find selected coefficients, including in a product of expressions.

Factorials and combinations

Why is the coefficient of $x^2$ in $(1+x)^5$ equal to $10$ ? Each of the five brackets offers a choice: take its $1$ or take its $x$ . To make $x^2$ , take $x$ from exactly two brackets. The coefficient counts how many different pairs of brackets you can choose.

There are $5\times4$ ways to choose a first bracket and then a second. This counts each pair twice, because choosing brackets 2 then 4 produces the same product as choosing 4 then 2. Dividing by $2$ gives $10$ .

Factorial

For a positive integer $n$ , factorial means

n!=n(n-1)(n-2)\cdots2\times1.

Also define $0!=1$ . For example, $4!=24$ ; it does not mean $4\times3$ .

The notation $\binom nr$ , also written ${}^nC_r$ , counts selections of $r$ objects from $n$ distinct objects when their order does not matter. Here, the objects are the brackets from which we choose the second term.

Binomial coefficients

For integers $n\ge0$ and $0\le r\le n$ ,

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

The two end coefficients are $\binom n0=\binom nn=1$ : there is one way to choose no brackets or all of them.

In the formula booklet.

Cancel factorials before multiplying. This keeps the calculation short and shows what is being counted.

Worked example 1: choosing two brackets

Question. Calculate $\binom52$ and explain why it is the coefficient of $x^2$ in $(1+x)^5$ .

Plan. Count selections of the two brackets supplying $x$ .

\binom52=\frac{5!}{2!3!} =\frac{5\times4}{2\times1}=10.

Each selection produces $1^3x^2=x^2$ , so ten such products combine.

Answer. $\binom52=10$ ; the coefficient is $10$ .

Check. Label the brackets 1 to 5. Grouping pairs by their smaller label gives $4+3+2+1=10$ pairs, a separate count.

Add one bracket, while still choosing two. The exponent of $x$ stays the same; the number of available choices changes.

Worked example 2: one more available bracket

Question. Calculate $\binom62$ and hence state the coefficient of $x^2$ in $(1+x)^6$ .

Plan. Use the same selection count with six brackets.

\binom62=\frac{6!}{2!4!} =\frac{6\times5}{2}=15.

Every pair still contributes one $x^2$ .

Answer. $\binom62=15$ , so the coefficient is $15$ .

Check. The old five brackets give $10$ pairs. The new bracket can be paired with each of those five, giving $10+5=15$ .

Factorials describe complete products down to $1$ .

Watch out

$\binom62\ne6!/2!$ . The denominator also needs $4!$ ; otherwise arrangements of the four unchosen brackets are counted repeatedly. Use $6!/(2!4!)$ .

Now change the number of brackets again.

The binomial expansion

The same counting works when a bracket contains $a+b$ . Choosing $b$ from $r$ brackets leaves $a$ from the other $n-r$ , so the product is $a^{n-r}b^r$ . There are $\binom nr$ ways to make it.

For example, choosing $x$ from two of four brackets in $(2+x)^4$ contributes $\binom42 2^2x^2=24x^2$ . The coefficient is not just $\binom42$ : the remaining brackets also contribute their factors of $2$ .

Positive integer binomial expansion

For a positive integer $n$ ,

(a+b)^n=a^n+\binom n1a^{n-1}b+\binom n2a^{n-2}b^2+\cdots+b^n.

The term obtained by choosing $b$ exactly $r$ times is $\binom nr a^{n-r}b^r$ . There are $n+1$ terms before any like terms are collected.

In the formula booklet.

Move through the expansion by reducing the power of the first term and increasing the power of the second. Their powers add to $n$ in every term. For a positive integer power this is a finite identity: it holds for every real value of the variable.

Worked example 3: a complete expansion

Question. Expand $(2+x)^4$ in ascending powers of $x$ .

Plan. Use coefficients $1,4,6,4,1$ , retaining the powers of $2$ .

\begin{aligned} (2+x)^4 &=2^4+4(2^3)x+6(2^2)x^2+4(2)x^3+x^4\\ &=16+32x+24x^2+8x^3+x^4. \end{aligned}

Answer. $16+32x+24x^2+8x^3+x^4$ .

Check. At $x=1$ , the original is $3^4=81$ ; the expansion gives $16+32+24+8+1=81$ . Multiplying $(4+4x+x^2)^2$ gives the same coefficients independently.

Change only $x$ to $-x$ . Odd powers change sign; even powers do not.

Worked example 4: a subtraction inside the bracket

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

Plan. Set the second term to $-x$ , including its sign inside every power.

\begin{aligned} (2-x)^4 &=2^4+4(2^3)(-x)+6(2^2)(-x)^2\\ &\quad+4(2)(-x)^3+(-x)^4\\ &=16-32x+24x^2-8x^3+x^4. \end{aligned}

Answer. $16-32x+24x^2-8x^3+x^4$ .

Check. At $x=1$ , this gives $16-32+24-8+1=1=(2-1)^4$ . Replacing $x$ by $-x$ in the independently checked previous expansion also produces every sign above.

A coefficient inside a bracket is raised to a power along with its variable.

Watch out

$(kx)^3=k^3x^3$ , not $kx^3$ . For instance, $(-2x)^3=-8x^3$ . Put the complete second term in brackets before cubing it.

Use that sign pattern with a new exponent and coefficient.

Finding a chosen coefficient

A request for one coefficient rarely needs the whole expansion. Identify which choice of $r$ produces the required power, then calculate only that term. A coefficient is the multiplying number: the coefficient of $x^3$ in $-80x^3$ is $-80$ .

In a product, work backwards from the target power. To make $x^3$ in $(3+x)F(x)$ , there are exactly two routes: multiply the $x^3$ term of $F$ by $3$ , or its $x^2$ term by $x$ . Add both contributions.

Worked example 5: an unknown inside a bracket

Question. The coefficient of $x^3$ in $(1+px)^5$ is $-80$ . Find $p$ .

Plan. Choose $px$ from three brackets and equate its coefficient to $-80$ .

\begin{aligned} \text{coefficient of }x^3&=\binom53 1^2p^3=10p^3\\ 10p^3&=-80\\ p^3&=-8\\ p&=-2. \end{aligned}

The cube includes the sign of $p$ .

Answer. $p=-2$ .

Check. Multiplying $(1-2x)^2=1-4x+4x^2$ by $(1-2x)^3=1-6x+12x^2-8x^3$ , the cubic contributions are $-8-48-24=-80$ , as required.

Keep this bracket and the value of $p$ , but multiply by one extra linear factor. Now two terms contribute to the answer.

Worked example 6: a coefficient in a product

Question. Using $p=-2$ , find the coefficient of $x^3$ in $(3+x)(1+px)^5$ .

Plan. Find the quadratic and cubic coefficients of $(1-2x)^5$ , then combine the two routes to $x^3$ .

\begin{aligned} \text{coefficient of }x^2&=\binom52(-2)^2=40\\ \text{coefficient of }x^3&=\binom53(-2)^3=-80. \end{aligned}

The required terms are $3(-80x^3)$ and $x(40x^2)$ .

\text{coefficient of }x^3=3(-80)+40=-200.

Answer. $-200$ .

Check. First multiply $(3+x)(1-2x)=3-5x-2x^2$ . The coefficients of $x,x^2,x^3$ in $(1-2x)^4$ are $-8,24,-32$ . The cubic coefficient of the regrouped product is $3(-32)-5(24)-2(-8)=-200$ .

The opening question now has its full explanation: choosing two $x$ terms from five brackets gives $\binom52=10$ . More complicated coefficients use the same choice count, with the chosen terms' own powers included.

Problem-solving

List the power combinations before calculating a coefficient in a product. For a cubic term, possible pairs are $0+3$ , $1+2$ , $2+1$ and $3+0$ ; keep only those available in the factors.

Try a product with fresh numbers.

Quick Recap

$n!$ is the product down to $1$ , with $0!=1$ .
$\binom nr$ counts which $r$ brackets supply the second term.
In $(a+b)^n$ , the corresponding term is $\binom nr a^{n-r}b^r$ .
Keep a signed term inside brackets when raising it to a power, and add every contribution to a requested coefficient in a product.

There are ten ways to choose the two brackets supplying $x$ in $(1+x)^5$ , so its quadratic coefficient is $10$ .

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1.6 - Expanding integer powers

Factorials and combinations

The binomial expansion

Finding a chosen coefficient

Quick Recap

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