Beyond expanding: exam questions often hide an unknown constant inside the bracket, or combine two expansions together, and ask you to work backwards from a coefficient.
11 min read4 worked examples
Beyond a straightforward expansion
So far, every expansion has had known numbers throughout. Two harder question types build on the same general-term method:
Unknown constants: the bracket contains a letter, such as $(1+ax)^n$, and you're told the value of one coefficient — the task is to find $a$ (or $n$). Combined expansions: two brackets are multiplied together, such as $(2+x)(1-3x)^5$, and you need a coefficient in the product — which means combining terms from a partial expansion of each bracket.
Strategy — isolate exactly the terms you need
Both problem types reward the same discipline as before: don't expand more than necessary. For an unknown-constant problem, write the general term and substitute the known power of $x$ to isolate the unknown. For a combined-expansion problem, work out which pairs of terms from each bracket multiply together to give the power of $x$ you want — usually only two or three terms from each bracket are relevant.
Worked example 1
Finding an unknown constant from a coefficient
In the expansion of $(1+ax)^6$, the coefficient of $x^3$ is $160$. Given that $a>0$, find the value of $a$.
MethodWrite the general term, then solve for a
1
General term: $\dbinom{6}{r}(ax)^r$, since $a=1$ for the first bracket entry, powers of $1$ vanish.
Only the $(ax)^r$ part carries the unknown.
2
We want $x^3$, so $r=3$: term is $\dbinom{6}{3}(ax)^3 = 20\,a^3x^3$.
$\binom{6}{3}=20$.
3
Set the coefficient equal to $160$: $20a^3 = 160$.
Match to the given information.
4
Solve: $a^3 = 8$, so $a=2$.
Cube root of 8 is 2; a>0 confirms the positive root.
▪ solved
Examiner's eye
The whole term $(ax)^3$ must be cubed, giving $a^3x^3$ — not $ax^3$. This is the same "bracket the whole term" discipline from ordinary expansions, just with a letter instead of a number.
Worked example 2
A coefficient from a product of two expansions
Find the coefficient of $x^2$ in the expansion of $(2+x)(1-3x)^4$.
Only a partial expansion of $(1-3x)^4$ is needed, up to $x^2$, since multiplying by $(2+x)$ can only raise the power by at most one more.
MethodExpand (1-3x)^4 up to x², then combine with (2+x)
1
Expand $(1-3x)^4$ up to the $x^2$ term: $\dbinom{4}{0}+\dbinom{4}{1}(-3x)+\dbinom{4}{2}(-3x)^2 = 1 - 12x + 54x^2$.
$\binom{4}{1}=4,\ \binom{4}{2}=6$; $(-3x)^2=9x^2$, so $6\times9=54$.
2
The $x^2$ term in the product $(2+x)(1-12x+54x^2+\cdots)$ comes from two pairings: $2 \times 54x^2$, and $x \times (-12x)$.
Any term from (2+x) times any term from the expansion that together give $x^2$.
3
$2\times54x^2 = 108x^2$, and $x\times(-12x) = -12x^2$.
Multiply each matching pair.
4
Add the contributions: $108x^2 - 12x^2 = 96x^2$.
Combine like terms.
5
The coefficient of $x^2$ is $96$.
Final answer.
▪ solved
The key insight
A coefficient in a product comes from every pair of terms whose powers add to the one you want — here, $x^0\times x^2$ and $x^1\times x^1$. Missing one pairing is the most common error in this question type; systematically list which powers from each bracket could combine before multiplying.
Worked example 3
An equation from a ratio of coefficients
In the expansion of $(1+x)^n$, the coefficient of $x^3$ is four times the coefficient of $x^2$. Find $n$.
MethodWrite both coefficients using nCr, then form an equation
1
Coefficient of $x^2$: $\dbinom{n}{2} = \dfrac{n(n-1)}{2}$. Coefficient of $x^3$: $\dbinom{n}{3}=\dfrac{n(n-1)(n-2)}{6}$.
Standard simplified forms.
2
Set up the given relationship: $\dfrac{n(n-1)(n-2)}{6} = 4\times\dfrac{n(n-1)}{2}$.
"Four times" translates to a factor of 4.
3
Since $n>2$, both sides contain the nonzero factor $n(n-1)$; divide it out: $\dfrac{n-2}{6} = \dfrac{4}{2} = 2$.
Simplifies the equation enormously.
4
Solve: $n-2 = 12$, so $n=14$.
Multiply both sides by 6.
▪ solved
Common mistake
Writing out $\binom{n}{2}$ and $\binom{n}{3}$ in full and cross-multiplying before simplifying leads to a messy cubic. Spotting the shared factor $n(n-1)$ and cancelling it first turns the problem into simple linear algebra — always look for a common factor before expanding.
Worked example 4
Two unknowns from two conditions · Challenge
In the expansion of $(1+ax)^n$, the coefficient of $x$ is $24$ and the coefficient of $x^2$ is $252$. Given that $a$ and $n$ are positive, find their values.
Two unknowns need two equations. The coefficients of $x$ and $x^2$ each give an equation in $a$ and $n$; dividing one by the other eliminates $a$ first.
MethodForm two equations, then eliminate
1
Coefficient of $x$: $\dbinom{n}{1}a = na$. So $na = 24$.
The $r=1$ term of the general expansion.
2
Coefficient of $x^2$: $\dbinom{n}{2}a^2 = \dfrac{n(n-1)}{2}a^2$. So $\dfrac{n(n-1)}{2}a^2 = 252$.
The $r=2$ term.
3
Divide the second equation by the first: $\dfrac{\frac{n(n-1)}{2}a^2}{na} = \dfrac{252}{24}$, which simplifies to $\dfrac{(n-1)a}{2} = 10.5$.
The n and one factor of a cancel.
4
So $(n-1)a = 21$. Combined with $na=24$, subtract: $na - (n-1)a = 24-21 \Rightarrow a = 3$.
Two linear equations in a and na eliminate cleanly.
5
Substitute back into $na=24$: $3n=24$, so $n=8$.
Solve for the remaining unknown.
▪ solved
Look deeper — dividing equations to eliminate a variable
Dividing one equation by another is a powerful move whenever both sides share a common factor you want to remove — here it stripped away most of the $n$-dependence in one step, leaving simple linear equations. It's a technique worth remembering beyond this topic, anywhere two related expressions are set up as separate equations.
Common mistake
When a bracket contains an unknown constant like $(1+ax)^n$, forgetting to raise $a$ to the same power as $x$ is the single most common slip — the term is $\binom{n}{r}a^rx^r$, with $a$ and $x$ both carrying the power $r$, not just $x$.
Check yourself
In the expansion of $(1+bx)^5$, the coefficient of $x^2$ is $270$. Given $b>0$, find $b$. (Hint: write the $r=2$ term of the general expansion and set it equal to 270.)