Partial Fractions

Partial fractions is the reverse of adding algebraic fractions. When you compute $\frac{2}{x-1} + \frac{3}{x+2}$ , you obtain a single fraction $\frac{5x+1}{(x-1)(x+2)}$ with a more complicated numerator and a factorised denominator. Partial-fraction decomposition runs the film backwards: starting from $\frac{5x+1}{(x-1)(x+2)}$ , it rebuilds the simpler summands $\frac{2}{x-1}$ and $\frac{3}{x+2}$ . The technique looks like a bookkeeping exercise, but its real importance is downstream. In the Edexcel A-Level, partial fractions unlock two otherwise intractable problems: integrating rational functions (Paper 2, section 7) and binomially expanding rational expressions of the form $(1+ax)^{-1}(1+bx)^{-1}$ as a power series (section 4). At university the same idea reappears in inverse Laplace transforms, in the residue calculus of complex analysis, and in the analysis of linear time-invariant systems in control engineering. Mastering the algebraic decomposition now pays compound interest later.

Spec mapping

This lesson is aligned with the Edexcel 9MA0 specification, Pure Mathematics Paper 1 — Section 2.10: "Decompose rational functions into partial fractions (denominators not more complicated than squared linear terms and with no more than 3 terms, numerators constant or linear)." The technique is then re-used in:

Paper 1, Section 4 — binomial expansion of $(1+x)^n$ for negative or fractional $n$ , applied to expressions such as $\frac{3x+1}{(1-x)(1+2x)}$ ;
Paper 2, Section 7 — integration of rational functions, where each partial fraction integrates to a logarithm or a reciprocal power;
Paper 2, Section 8 — first-order separable differential equations in which the variable separates into a rational function of $x$ .

Edexcel restrict denominators to (at most) a product of three factors, each linear or a single irreducible quadratic, with numerators of degree at most one. The formula booklet does not list partial-fraction templates — you must memorise the three standard forms.

When does partial fractions apply?

Partial fractions decomposes a proper rational function — a quotient $\frac{P(x)}{Q(x)}$ in which $\deg P < \deg Q$ . If $\deg P \geq \deg Q$ the fraction is improper and you must polynomial-long-divide first, leaving a polynomial quotient plus a proper remainder fraction; only the remainder is decomposed.

A theorem from algebra (sometimes called the partial-fraction decomposition theorem for rational functions over $\mathbb{R}[x]$ ) guarantees that every proper rational function with real coefficients can be written uniquely as a sum of fractions whose denominators are powers of the irreducible factors of $Q(x)$ . Over the reals, the irreducible factors are exactly the linear polynomials $(ax+b)$ and the irreducible quadratics $(ax^2+bx+c)$ with negative discriminant. The Edexcel syllabus restricts you to three standard cases.

The three Edexcel cases

Case	Denominator factor	Partial-fraction form
(a) Distinct linear	$(ax+b)$ appearing once	$\dfrac{A}{ax+b}$
(b) Repeated linear	$(ax+b)^2$	$\dfrac{A}{ax+b} + \dfrac{B}{(ax+b)^2}$
(c) Irreducible quadratic	$(ax^2+bx+c)$ with $b^2-4ac<0$	$\dfrac{Ax+B}{ax^2+bx+c}$

You combine these case-by-case for each factor of $Q(x)$ . For example, $\frac{P(x)}{(x-1)(x+2)^2(x^2+1)}$ decomposes as

$\frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2} + \frac{Dx+E}{x^2+1}.$

The number of unknowns $A, B, C, D, E$ matches the degree of $Q(x)$ — here, degree 5 = 1 + 1 + 1 + 2.

Three solution methods

There are three techniques for finding the unknown numerators, and you should be fluent in all three because each is fastest in different settings.

Method	Best for	Mechanism
Cover-up (Heaviside)	Distinct linear factors	Multiply by the factor, substitute the value that kills it
Substitution of $x$	Distinct linear, repeated linear	Multiply through by $Q(x)$ , then substitute clever values of $x$
Comparing coefficients	Repeated linear, irreducible quadratic	Multiply through by $Q(x)$ , expand, equate powers of $x$

In practice, use cover-up for any distinct linear factor (one second of work), substitution for the "easy" constants in repeated-linear cases, and comparing coefficients to mop up the remaining unknowns when substitution leaves them entangled.

The cover-up method (Heaviside)

The cover-up method, attributed to Oliver Heaviside in the late nineteenth century during his work on operational calculus for telegraph signals, is the fastest way to extract the numerator above a distinct linear factor.

Recipe. To find $A$ in $\frac{P(x)}{(x-\alpha)Q_1(x)} = \frac{A}{x-\alpha} + \cdots$ , cover up the $(x-\alpha)$ factor in the original denominator and substitute $x=\alpha$ into what remains:

$A = \left.\frac{P(x)}{Q_1(x)}\right|_{x=\alpha}.$

It works because multiplying both sides by $(x-\alpha)$ and then setting $x=\alpha$ kills every other term on the right. The method extends to repeated factors by differentiation (see A vs A* divide below) but is most efficient as stated for the distinct-linear case.

The comparing-coefficients method in detail

The comparing-coefficients method is the most general technique. After multiplying through by $Q(x)$ to clear denominators, you have a polynomial identity that must hold for all $x$ . Two polynomials agree at every $x$ iff their coefficients agree power-by-power, so collecting the coefficient of $x^k$ on each side gives one equation per power. The total number of equations equals the degree of the cleared identity, which equals the number of unknowns — the system is square and (by uniqueness of decomposition) non-singular.

Worked illustration. Decompose $\dfrac{3x^2 + 4x - 5}{(x-1)(x^2 + 2)}$ where $x^2+2$ is irreducible. The template is

$\frac{3x^2 + 4x - 5}{(x-1)(x^2 + 2)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+2}.$

Multiply through:

$3x^2 + 4x - 5 = A(x^2+2) + (Bx+C)(x-1).$

Full-substitution route. Substitute $x=1$ to kill the second term: $3 + 4 - 5 = 3A \Rightarrow A = \tfrac{2}{3}$ . Substitute $x=0$ : $-5 = 2A - C \Rightarrow C = 2A + 5 = \tfrac{4}{3} + 5 = \tfrac{19}{3}$ . Substitute $x=-1$ : $3 - 4 - 5 = 3A + (-B+C)(-2) = 2 + 2(B-C) \Rightarrow -6 = 2 + 2B - \tfrac{38}{3} \Rightarrow 2B = -8 + \tfrac{38}{3} = \tfrac{14}{3}$ , so $B = \tfrac{7}{3}$ .

Matrix-form (collect-coefficients) route. Expand the right-hand side:

$A(x^2+2) + (Bx+C)(x-1) = (A+B)x^2 + (-B+C)x + (2A - C).$

Equate coefficients power-by-power against $3x^2 + 4x - 5$ to obtain three simultaneous equations:

$\begin{aligned} x^2: \quad & A + B = 3 \\ x^1: \quad & -B + C = 4 \\ x^0: \quad & 2A - C = -5 \end{aligned}$

In matrix form $\begin{pmatrix} 1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 0 & -1 \end{pmatrix} \begin{pmatrix} A \\ B \\ C \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix}$ . Add row 3 to twice row 1: $4A + 2B = 1$ combined with $A+B=3 \Rightarrow 2A = -5 \Rightarrow A = \tfrac{2}{3}$ — actually solving cleanly: from row 1, $B = 3 - A$ ; substituting in row 2 gives $C = 4 + B = 7 - A$ ; row 3 then gives $2A - (7-A) = -5 \Rightarrow 3A = 2 \Rightarrow A = \tfrac{2}{3}$ , then $B = \tfrac{7}{3}$ , $C = \tfrac{19}{3}$ . Both routes agree.

The matrix view is conceptually cleaner: it shows partial fractions as a linear problem in the unknowns, and it generalises to computer-algebra implementations. The substitution route is faster by hand for small systems. A* candidates are fluent in both and choose by problem size.

Worked example 1 — Distinct linear factors

Decompose $\dfrac{5x+1}{(x-1)(x+2)}$ into partial fractions.

Step 1 — write the form. Two distinct linear factors give

$\frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}.$

Step 2 — apply cover-up for $A$ . Cover the $(x-1)$ in the denominator and substitute $x=1$ :

$A = \frac{5(1)+1}{(1)+2} = \frac{6}{3} = 2.$

Step 3 — apply cover-up for $B$ . Cover the $(x+2)$ and substitute $x=-2$ :

$B = \frac{5(-2)+1}{(-2)-1} = \frac{-9}{-3} = 3.$

Step 4 — state the answer.

$\frac{5x+1}{(x-1)(x+2)} = \frac{2}{x-1} + \frac{3}{x+2}.$

Step 5 — check by recombining. $\frac{2(x+2) + 3(x-1)}{(x-1)(x+2)} = \frac{5x+1}{(x-1)(x+2)}$ . Correct.

Worked example 2 — Repeated linear factor

Decompose $\dfrac{4x+5}{(x+1)(x-2)^2}$ into partial fractions.

Step 1 — write the form. A distinct linear $(x+1)$ and a repeated linear $(x-2)^2$ give

$\frac{4x+5}{(x+1)(x-2)^2} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}.$

Step 2 — multiply through by the common denominator.

$4x+5 = A(x-2)^2 + B(x+1)(x-2) + C(x+1). \quad (\star)$

Step 3 — substitute $x=-1$ (kills the $A$ -term... wait, the others). Setting $x=-1$ kills $B$ and $C$ :

$4(-1)+5 = A(-3)^2 \implies 1 = 9A \implies A = \tfrac{1}{9}.$

Step 4 — substitute $x=2$ . This kills $A$ and $B$ :

$4(2)+5 = C(3) \implies 13 = 3C \implies C = \tfrac{13}{3}.$

Step 5 — find $B$ by comparing coefficients. Expand $(\star)$ :

$A(x-2)^2 = A(x^2 - 4x + 4),$ $B(x+1)(x-2) = B(x^2 - x - 2),$ $C(x+1) = Cx + C.$

Coefficient of $x^2$ : $A + B = 0$ , so $B = -A = -\tfrac{1}{9}$ .

(One could equally substitute any further value of $x$ , e.g. $x=0$ , and solve for $B$ .)

Step 6 — state the answer.

$\frac{4x+5}{(x+1)(x-2)^2} = \frac{1/9}{x+1} - \frac{1/9}{x-2} + \frac{13/3}{(x-2)^2}.$

Worked example 3 — Irreducible quadratic factor

Decompose $\dfrac{2x^2 - x + 4}{(x+1)(x^2+4)}$ into partial fractions.

Step 1 — confirm $(x^2+4)$ is irreducible. Discriminant of $x^2 + 4$ is $0 - 16 = -16 < 0$ , so it has no real linear factors. Use a linear numerator:

$\frac{2x^2 - x + 4}{(x+1)(x^2+4)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+4}.$

Step 2 — multiply through.

$2x^2 - x + 4 = A(x^2+4) + (Bx+C)(x+1).$

Step 3 — substitute $x=-1$ for $A$ .

$2(1) - (-1) + 4 = A(1+4) \implies 7 = 5A \implies A = \tfrac{7}{5}.$

Step 4 — compare coefficients of $x^2$ and constants.

Coefficient of $x^2$ : $A + B = 2 \implies B = 2 - \tfrac{7}{5} = \tfrac{3}{5}.$

Constant term: $4A + C = 4 \implies C = 4 - \tfrac{28}{5} = -\tfrac{8}{5}.$

Step 5 — state the answer.

$\frac{2x^2 - x + 4}{(x+1)(x^2+4)} = \frac{7/5}{x+1} + \frac{(3/5)x - 8/5}{x^2+4}.$

Sidebar — why does the numerator over an irreducible quadratic have to be linear?

Count the degrees of freedom. A proper rational function $P(x)/Q(x)$ with $\deg Q = n$ has a numerator $P(x)$ of degree at most $n-1$ , so it is determined by $n$ free coefficients. The decomposition must therefore also have exactly $n$ unknowns — no more, no less, or the matching to $P(x)$ would be over- or under-determined.

Each linear factor $(x-\alpha)$ contributes one unknown (the constant $A$ above $\frac{A}{x-\alpha}$ ). Each repeated linear factor $(x-\alpha)^k$ contributes $k$ unknowns ( $A_1, \ldots, A_k$ ). Each irreducible quadratic $(x^2+bx+c)$ contributes two unknowns (the $A$ and $B$ in $\frac{Ax+B}{x^2+bx+c}$ ) — exactly because the factor itself has degree 2.

Add up: $\sum (\text{contribution from each factor}) = \deg Q = n$ , matching the $n$ free coefficients of $P(x)$ . The decomposition has exactly the right degrees of freedom to represent any proper numerator. If you mistakenly wrote $\frac{A}{x^2+bx+c}$ (one unknown for a quadratic factor), you would be one short, and the system $A(\cdots) + \cdots = P(x)$ would generically have no solutions.

This counting argument is a corollary of the Fundamental Theorem of Algebra (Gauss, 1799): every polynomial of degree $n$ over $\mathbb{C}$ has exactly $n$ roots counted with multiplicity. Over $\mathbb{R}$ , complex roots come in conjugate pairs, so each pair bundles into a single real irreducible quadratic — and the "two" in " $Ax+B$ " is exactly the count of those bundled complex roots.

Worked example 4 — Improper rational function

Decompose $\dfrac{x^3 + 2x^2 - x + 1}{x^2 - 1}$ into partial fractions.

Step 1 — recognise it is improper. $\deg(\text{numerator}) = 3 > 2 = \deg(\text{denominator})$ , so polynomial long division is required first.

Step 2 — divide.

$x^3 + 2x^2 - x + 1 = (x^2 - 1)(x + 2) + (3).$

(Check: $(x^2-1)(x+2) = x^3 + 2x^2 - x - 2$ , so the remainder is $x^3 + 2x^2 - x + 1 - (x^3 + 2x^2 - x - 2) = 3$ .)

Hence

$\frac{x^3 + 2x^2 - x + 1}{x^2-1} = x + 2 + \frac{3}{(x-1)(x+1)}.$

Step 3 — decompose the proper remainder.

$\frac{3}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}.$

Cover-up: $A = \tfrac{3}{1+1} = \tfrac{3}{2}$ , $B = \tfrac{3}{-1-1} = -\tfrac{3}{2}$ .

Step 4 — state the full answer.

$\frac{x^3 + 2x^2 - x + 1}{x^2-1} = x + 2 + \frac{3/2}{x-1} - \frac{3/2}{x+1}.$

Why a linear quotient appeared. When polynomial-dividing $P(x)$ by $Q(x)$ , the quotient has degree $\deg P - \deg Q$ (provided the division terminates, which it always does over $\mathbb{R}[x]$ ). Here $\deg P = 3$ and $\deg Q = 2$ , so the quotient is degree $3 - 2 = 1$ — a linear $x + 2$ . If the numerator had been degree 4 the quotient would be quadratic; degree 5, cubic. The remainder always has degree strictly less than $\deg Q$ (otherwise division could continue), so the proper-remainder fraction is guaranteed decomposable by the standard cases. This is why the first check on any rational function is to compare degrees: it tells you both whether long division is needed and what shape the polynomial part will take.

Worked example 4b — A more elaborate improper case

Decompose $\dfrac{2x^4 - x^3 + 5x - 1}{(x-1)(x+2)}$ into partial fractions.

Step 1 — degree check. Numerator degree 4, denominator degree 2, so the quotient will have degree $4 - 2 = 2$ (a quadratic). Long division is required.

Step 2 — multiply out the denominator. $(x-1)(x+2) = x^2 + x - 2$ .

Step 3 — long divide $2x^4 - x^3 + 5x - 1$ by $x^2 + x - 2$ .

$2x^4 \div x^2 = 2x^2$ ; subtract $2x^2(x^2+x-2) = 2x^4 + 2x^3 - 4x^2$ to leave $-3x^3 + 4x^2 + 5x - 1$ .
$-3x^3 \div x^2 = -3x$ ; subtract $-3x(x^2+x-2) = -3x^3 - 3x^2 + 6x$ to leave $7x^2 - x - 1$ .
$7x^2 \div x^2 = 7$ ; subtract $7(x^2+x-2) = 7x^2 + 7x - 14$ to leave $-8x + 13$ .

So $\dfrac{2x^4 - x^3 + 5x - 1}{(x-1)(x+2)} = 2x^2 - 3x + 7 + \dfrac{-8x + 13}{(x-1)(x+2)}$ .

Step 4 — decompose the proper remainder. Cover-up at $x=1$ : $A = \dfrac{-8(1)+13}{1+2} = \dfrac{5}{3}$ . Cover-up at $x=-2$ : $B = \dfrac{-8(-2)+13}{-2-1} = \dfrac{29}{-3} = -\dfrac{29}{3}$ .

Step 5 — state the answer.

$\frac{2x^4 - x^3 + 5x - 1}{(x-1)(x+2)} = 2x^2 - 3x + 7 + \frac{5/3}{x-1} - \frac{29/3}{x+2}.$

The polynomial part $2x^2 - 3x + 7$ dominates the function for large $|x|$ ; the partial-fraction part dominates near the poles at $x = 1$ and $x = -2$ . This decomposition is exactly what is needed to integrate the original function — the polynomial integrates termwise, and each partial fraction integrates to a logarithm.

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (7 marks):

(a) Express $f(x) = \dfrac{9x^2 + 8x - 1}{(1-x)(1+2x)^2}$ in partial fractions. (4)

(b) Hence, or otherwise, find the binomial expansion of $f(x)$ in ascending powers of $x$ , up to and including the term in $x^2$ . (3)

Solution with mark scheme:

(a) Form. $\dfrac{9x^2 + 8x - 1}{(1-x)(1+2x)^2} = \dfrac{A}{1-x} + \dfrac{B}{1+2x} + \dfrac{C}{(1+2x)^2}.$

M1 (AO1.1a) — correct partial-fraction template, with $\dfrac{B}{1+2x} + \dfrac{C}{(1+2x)^2}$ for the repeated factor (not just one term).

Multiply through:

$9x^2 + 8x - 1 = A(1+2x)^2 + B(1-x)(1+2x) + C(1-x).$

Substitute $x=1$ : $9 + 8 - 1 = A(3)^2 \implies 16 = 9A \implies A = \tfrac{16}{9}$ .

Substitute $x=-\tfrac{1}{2}$ : $9(\tfrac{1}{4}) + 8(-\tfrac{1}{2}) - 1 = C(\tfrac{3}{2}) \implies \tfrac{9}{4} - 4 - 1 = \tfrac{3}{2}C \implies -\tfrac{19}{4} = \tfrac{3}{2}C \implies C = -\tfrac{19}{6}$ .

Partial Fractions

Partial Fractions

Spec mapping

When does partial fractions apply?

The three Edexcel cases

Three solution methods

The cover-up method (Heaviside)

The comparing-coefficients method in detail

Worked example 1 — Distinct linear factors

Worked example 2 — Repeated linear factor

Worked example 3 — Irreducible quadratic factor

Sidebar — why does the numerator over an irreducible quadratic have to be linear?

Worked example 4 — Improper rational function

Worked example 4b — A more elaborate improper case

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

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