Some rational expressions have no binomial form of their own to exploit — split them into partial fractions first, and two familiar expansions fall out instead.
10 min read3 worked examples
Why some expressions need splitting first
A rational function like $\dfrac{5-x}{(1-x)(2+x)}$ doesn't match either $(1+x)^n$ or $(a+bx)^n$ directly — its denominator is a product of two linear factors, not a single bracket raised to a power. There is no way to factor out a single constant and land on the form $a^n(1+u)^n$. The fix is to decompose the expression into partial fractions first, splitting it into a sum of two (or more) simple fractions, each of which does match a binomial form once rewritten as $\frac{k}{1\pm cx}=k(1\pm cx)^{-1}$.
Each partial fraction is then expanded separately using the techniques from the previous two lessons, and the results are added together term by term.
Combining validity ranges
Each partial fraction generally has its own condition for convergence, since each has a different denominator. The combined expansion is only valid where every individual expansion is valid simultaneously — so the overall range of validity is the intersection (the more restrictive) of all the individual ranges, never the union.
Worked example 1
Combining with partial fractions
Given that $\dfrac{5-x}{(1-x)(2+x)} \equiv \dfrac{2}{1-x}+\dfrac{-1}{2+x}$, find the binomial expansion of $\dfrac{5-x}{(1-x)(2+x)}$ up to and including the term in $x^2$, and state the range of values of $x$ for which the expansion is valid.
MethodExpand each partial fraction separately, then combine and intersect the validity ranges
1
$\dfrac{2}{1-x} = 2(1-x)^{-1} = 2\big[1+(-1)(-x)+\tfrac{(-1)(-2)}{2}(-x)^2+\cdots\big] = 2\big[1+x+x^2+\cdots\big]$, valid for $|x|<1$.
Expand the first partial fraction as $(1+u)^{-1}$ with $u=-x$; since $|u|=|x|$, validity is $|x|<1$ directly.
2
$\dfrac{-1}{2+x} = -\dfrac12\left(1+\dfrac{x}{2}\right)^{-1} = -\dfrac12\big[1-\tfrac{x}{2}+\tfrac{3x^2}{4}+\cdots\big]$, valid for $\left|\dfrac{x}{2}\right|<1 \Rightarrow |x|<2$.
Factor out 2 first, expand with $u=\frac{x}{2}$, $n=-1$, then apply validity to $u$.
Combine the two expansions term by term, up to $x^2$.
4
$= \dfrac32 + \dfrac94x + \dfrac{13}{8}x^2$.
Simplify each coefficient.
5
The combined expansion is only valid where both individual expansions are valid: $|x|<1$ and $|x|<2$ together give $|x|<1$.
The overall range of validity is the intersection (the more restrictive) of the two individual ranges.
▪ solved
Look deeper — why partial fractions unlock this expansion
$\dfrac{5-x}{(1-x)(2+x)}$ has no direct binomial form of its own — but splitting it into partial fractions turns it into a sum of two expressions that do match $(1+u)^n$ after factoring. This is precisely why partial fractions and the binomial expansion are often tested together: the decomposition is the bridge that makes an otherwise inaccessible expression expandable.
Worked example 2
A different partial-fraction decomposition
Given that $\dfrac{4+5x}{(1-x)(1+2x)} \equiv \dfrac{3}{1-x}+\dfrac{1}{1+2x}$, find the binomial expansion of $\dfrac{4+5x}{(1-x)(1+2x)}$ up to and including the term in $x^2$, and state the range of values of $x$ for which the expansion is valid.
MethodExpand each partial fraction separately, then combine and intersect the validity ranges
1
$\dfrac{3}{1-x} = 3(1-x)^{-1} = 3\big[1+x+x^2+\cdots\big] = 3+3x+3x^2$, valid for $|x|<1$.
Expand as $(1+u)^{-1}$ with $u=-x$; since $|u|=|x|$, validity is $|x|<1$ directly.
Expand as $(1+u)^{-1}$ with $u=2x$, no factoring needed since the leading term is already 1.
3
Adding: $(3+3x+3x^2) + (1-2x+4x^2) = 4+x+7x^2$.
Combine the two expansions term by term, up to $x^2$.
4
The combined expansion is valid where both hold: $|x|<1$ and $|x|<\dfrac12$ together give $|x|<\dfrac12$.
The more restrictive of the two ranges determines the overall validity.
▪ solved
Worked example 3
Finding the partial fractions before expanding · Challenge
Express $\dfrac{7-x}{(1+2x)(1-x)}$ in partial fractions, then find its binomial expansion up to and including the term in $x^2$, stating the range of values of $x$ for which the expansion is valid.
MethodDerive the partial fractions first, then expand each one and combine
1
Let $\dfrac{7-x}{(1+2x)(1-x)} \equiv \dfrac{A}{1+2x}+\dfrac{B}{1-x}$. Multiplying through: $7-x \equiv A(1-x)+B(1+2x)$.
Standard partial-fractions setup for two distinct linear factors.
2
Let $x=1$: $6 = 3B \Rightarrow B=2$. Let $x=-\dfrac12$: $\dfrac{15}{2} = \dfrac32A \Rightarrow A=5$.
Substitute the values of $x$ that each make one bracket vanish.
The partial-fraction decomposition, ready to expand.
4
$\dfrac{5}{1+2x} = 5(1+2x)^{-1} = 5\big[1-2x+4x^2+\cdots\big] = 5-10x+20x^2$, valid for $|x|<\dfrac12$.
Expand with $u=2x$; apply validity to $u$.
5
$\dfrac{2}{1-x} = 2(1-x)^{-1} = 2\big[1+x+x^2+\cdots\big] = 2+2x+2x^2$, valid for $|x|<1$.
Expand with $u=-x$; since $|u|=|x|$, validity is $|x|<1$ directly.
6
Adding: $(5-10x+20x^2)+(2+2x+2x^2) = 7-8x+22x^2$.
Combine the two expansions term by term, up to $x^2$.
7
Combined validity: $|x|<\dfrac12$ and $|x|<1$ together give $|x|<\dfrac12$.
The more restrictive range wins.
▪ solved
Look deeper
Unlike Examples 1 and 2, the partial-fraction identity here isn't handed to you — it has to be derived first, using exactly the same cover-up or comparing-coefficients technique from the Algebraic Methods topic. Deriving the decomposition and then expanding are two entirely separate skills, and a question can test either one, or both together, as it does here.
Common mistake
Combining the two partial fractions' validity ranges incorrectly — taking the less restrictive range (a union) instead of the more restrictive one (an intersection). The combined series can only be trusted where every individual piece it was built from is itself valid, so the smaller of the two ranges always wins.
Check yourself
Given that $\dfrac{5+3x}{(1-x)(1+3x)} \equiv \dfrac{2}{1-x}+\dfrac{3}{1+3x}$, find the binomial expansion of $\dfrac{5+3x}{(1-x)(1+3x)}$ up to and including the term in $x^2$, and state the range of values of $x$ for which the expansion is valid. (Hint: expand each partial fraction separately before combining.)
Method
1
$\dfrac{2}{1-x} = 2(1+x+x^2+\cdots) = 2+2x+2x^2$, valid for $|x|<1$.
Expand with $u=-x$.
2
$\dfrac{3}{1+3x} = 3(1-3x+9x^2+\cdots) = 3-9x+27x^2$, valid for $|3x|<1 \Rightarrow |x|<\dfrac13$.
Expand with $u=3x$; apply validity to $u$.
3
Adding: $(2+2x+2x^2)+(3-9x+27x^2) = 5-7x+29x^2$.
Combine term by term, up to $x^2$.
4
Combined validity: $|x|<1$ and $|x|<\dfrac13$ together give $|x|<\dfrac13$.