Further Partial Fractions

At A-Level Mathematics you decompose a proper algebraic fraction whose denominator factorises into distinct linear factors. Further Mathematics removes those two restrictions: the denominator may contain a repeated linear factor $(x-a)^n$ , an irreducible quadratic $x^2+bx+c$ with no real roots, or the fraction may be improper (numerator degree $\ge$ denominator degree). Each case demands a different form for the decomposition, and choosing that form correctly — before any algebra — is what separates a clean solution from a stalled one.

This lesson rebuilds the technique from the Cover-Up Rule upward, proves why each extra term is needed, and drills the three exam-critical cases with full mark schemes. Partial fractions are rarely an end in themselves: they are the gateway to integration ( $\int \tfrac{1}{x-a}\,dx = \ln|x-a|$ ), to the method of differences, and to Maclaurin/binomial series — so fluency here pays compound interest across the whole course.

Where this sits in AQA 7367

This is the partial fractions strand of the compulsory pure content examined on Paper 1 and Paper 2 (each 2 hours, 100 marks, 33⅓% of the A-Level). It promotes the A-Level Maths "distinct linear factors" technique to the repeated-factor, irreducible-quadratic and improper cases. The decomposition itself is largely AO1 (use a routine procedure), but the "hence integrate" or "hence find the sum of the series" follow-ons that examiners attach carry the AO2/AO3 marks — partial fractions almost never appear in isolation. The skill feeds directly into further calculus (integrating rational functions) and the method of differences lesson, where the telescoping split is a partial-fraction decomposition.

Core theory: matching the form to the denominator

A partial-fraction decomposition rests on one structural fact: a proper rational function is uniquely determined by a sum of "simplest" fractions, one family per factor of the denominator. The art is knowing which family each factor demands. The rule is governed entirely by the type and multiplicity of each factor.

Denominator factor	Term(s) it contributes
distinct linear $(ax+b)$	$\dfrac{A}{ax+b}$
repeated linear $(x-a)^n$	$\dfrac{A_1}{x-a} + \dfrac{A_2}{(x-a)^2} + \cdots + \dfrac{A_n}{(x-a)^n}$
irreducible quadratic $(x^2+bx+c)$	$\dfrac{Bx+C}{x^2+bx+c}$ (numerator one degree lower)

There is a single counting principle behind the whole table: the number of unknown constants must equal the degree of the denominator. A denominator $(x+1)^2(x-2)$ has degree $3$ , so the decomposition has three constants $A, B, C$ ; a denominator $(x-1)(x^2+1)$ also has degree $3$ ( $A$ plus $B, C$ from the quadratic). If your proposed form has the wrong number of constants, it is structurally wrong before you start — this is the fastest self-check in the topic.

Why a repeated factor needs every lower power

Students often ask why $(x-a)^2$ needs both $\tfrac{A_1}{x-a}$ and $\tfrac{A_2}{(x-a)^2}$ , not just the second term. The reason is a counting argument: a single fraction $\tfrac{A_2}{(x-a)^2}$ has only one degree of freedom, but to match an arbitrary linear numerator over $(x-a)^2$ you need two. Concretely, any expression $\tfrac{px+q}{(x-a)^2}$ can be rewritten by setting $px+q = p(x-a) + (pa+q)$ :

$\frac{px+q}{(x-a)^2} = \frac{p(x-a) + (pa+q)}{(x-a)^2} = \frac{p}{x-a} + \frac{pa+q}{(x-a)^2}.$

The first term $\tfrac{p}{x-a}$ appears unavoidably — drop it and you cannot represent a general linear numerator. The same argument iterates: $(x-a)^n$ needs all powers $1$ through $n$ , giving $n$ constants for a degree- $n$ factor.

Why an irreducible quadratic gets a linear numerator

If $x^2+bx+c$ has no real roots it cannot be split further over the reals, so it stays as a single denominator. But a constant numerator $\tfrac{A}{x^2+bx+c}$ again offers too little freedom: a degree-2 denominator can support a numerator of degree up to $1$ , so the correct numerator is the general linear $Bx+C$ . The "one degree lower than the denominator" rule is universal — it is why a linear factor gets a constant (degree 0) and a quadratic gets a linear (degree 1) numerator.

Two routes to the constants

Once the form is set and you have cleared denominators, two methods find the constants, and the best solutions combine them:

Substitution (the Cover-Up Rule). Substitute the root of each linear factor to kill all but one term. For a factor $(x-a)$ , setting $x=a$ isolates its constant instantly. This is fast and clean for linear factors but does not directly give the constants over an irreducible quadratic.
Comparing coefficients. Equate coefficients of $x^n$ on both sides of the cleared identity. Indispensable for the quadratic numerators $Bx+C$ and for the highest-power constant of a repeated factor. Comparing the highest power ( $x^2$ , $x^3$ , …) is often the quickest single equation.

The professional habit: use substitution to grab every constant the roots will give you, then compare one well-chosen coefficient (usually the highest power, or the constant term) to mop up the rest.

The Cover-Up Rule, made concrete

The Cover-Up Rule deserves a worked illustration because it is so fast when it applies. To decompose $\dfrac{px+q}{(x-a)(x-b)}$ into $\dfrac{A}{x-a} + \dfrac{B}{x-b}$ , you find $A$ by covering up the factor $(x-a)$ in the original fraction and evaluating what remains at $x=a$ : $A = \dfrac{pa+q}{a-b}$ . Mechanically, this works because multiplying the identity by $(x-a)$ and setting $x=a$ annihilates the $B$ -term, leaving $A$ equal to the "covered" fraction at $x=a$ . For example, in $\dfrac{5}{(x-1)(x+4)}$ , covering $(x-1)$ and putting $x=1$ gives $A = \dfrac{5}{1+4} = 1$ ; covering $(x+4)$ and putting $x=-4$ gives $B = \dfrac{5}{-4-1} = -1$ . Two constants in seconds, no simultaneous equations. The rule extends to the highest power of a repeated factor and to the linear factor sitting alongside a quadratic — but, crucially, it cannot reach the lower powers of a repeated factor or the $Bx+C$ of an irreducible quadratic, which is exactly why comparing coefficients remains indispensable.

Worked examples (with mark scheme)

Example 1 — a repeated linear factor

Express $\dfrac{3x + 5}{(x + 1)^2(x - 2)}$ in partial fractions.

Set up the form with the repeated factor giving two terms (three constants for the degree-3 denominator):

$\frac{3x + 5}{(x + 1)^2(x - 2)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x - 2}. \quad (\text{M1: correct form})$

Multiply through by $(x + 1)^2(x - 2)$ :

$3x + 5 = A(x + 1)(x - 2) + B(x - 2) + C(x + 1)^2. \quad (\text{M1: clear denominators})$

Substitute the roots to isolate constants:

$x = -1:\quad 3(-1)+5 = 2 = B(-1-2) = -3B \;\Rightarrow\; B = -\tfrac{2}{3}. \quad (\text{A1})$ $x = 2:\quad 3(2)+5 = 11 = C(2+1)^2 = 9C \;\Rightarrow\; C = \tfrac{11}{9}. \quad (\text{A1})$

The substitution $x=-1$ cannot give $A$ (its term vanishes too), so compare the $x^2$ coefficient. On the right, the $x^2$ terms are $A$ (from $A(x+1)(x-2)$ ) and $C$ (from $C(x+1)^2$ ); the left has no $x^2$ :

$0 = A + C \;\Rightarrow\; A = -C = -\tfrac{11}{9}. \quad (\text{M1 compare; A1})$

$\therefore\; \frac{3x + 5}{(x + 1)^2(x - 2)} = -\frac{11}{9(x + 1)} - \frac{2}{3(x + 1)^2} + \frac{11}{9(x - 2)}.$

(M1 form; M1 clear; A1 $B$ ; A1 $C$ ; M1 compare $x^2$ ; A1 $A$ . Check at $x=0$ : RHS $= -\tfrac{11}{9} - \tfrac{2}{3} + \tfrac{11}{9}\cdot(-\tfrac12) = -\tfrac{11}{9} - \tfrac{2}{3} - \tfrac{11}{18} = \tfrac{-22 - 12 - 11}{18} = -\tfrac{45}{18} = -\tfrac52$ ; LHS $= \tfrac{5}{(1)(-2)} = -\tfrac52$ ✓.)

Example 2 — an irreducible quadratic factor

Express $\dfrac{5x^2 + 3x + 1}{(x - 1)(x^2 + 1)}$ in partial fractions.

The quadratic $x^2+1$ has discriminant $0^2 - 4(1)(1) = -4 < 0$ , so it is irreducible and takes a linear numerator:

$\frac{5x^2 + 3x + 1}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}. \quad (\text{M1: correct form})$

Clear denominators:

$5x^2 + 3x + 1 = A(x^2 + 1) + (Bx + C)(x - 1). \quad (\text{M1})$

Substitute the linear root $x = 1$ : $5 + 3 + 1 = 9 = A(2)$ , so $A = \tfrac{9}{2}$ . (A1)

Compare $x^2$ : $5 = A + B$ , so $B = 5 - \tfrac{9}{2} = \tfrac{1}{2}$ . (A1)

Compare constants ( $x^0$ ): $1 = A - C$ , so $C = A - 1 = \tfrac{7}{2}$ . (A1)

$\therefore\; \frac{5x^2 + 3x + 1}{(x - 1)(x^2 + 1)} = \frac{9}{2(x - 1)} + \frac{x + 7}{2(x^2 + 1)}.$

(M1 form; M1 clear; A1 $A$ ; A1 $B$ via $x^2$ ; A1 $C$ via constants. As a check, compare the $x^1$ coefficient — an equation we did not use: RHS gives $-B + C = -\tfrac12 + \tfrac72 = 3$ , matching the LHS coefficient $3$ ✓. Using an unused coefficient to verify is the mark of a careful candidate.)

Example 3 — an improper fraction (divide first)

Express $\dfrac{x^3 + 1}{x^2 - 1}$ in the form $P(x) +$ partial fractions.

Numerator degree $3 \ge$ denominator degree $2$ : the fraction is improper, so divide first. By long division (or by writing $x^3 + 1 = x(x^2-1) + (x+1)$ ):

$\frac{x^3 + 1}{x^2 - 1} = x + \frac{x + 1}{x^2 - 1}. \quad (\text{M1: divide})$

Now the remainder is proper. Factorise $x^2 - 1 = (x-1)(x+1)$ and cancel the common $(x+1)$ :

$\frac{x + 1}{(x - 1)(x + 1)} = \frac{1}{x - 1}. \quad (\text{M1: simplify; A1})$

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

(M1 for the division reducing the fraction to proper; M1 for spotting the cancelling factor; A1 final. Spotting that $x+1$ cancels avoids the longer route of setting up $\tfrac{A}{x-1}+\tfrac{B}{x+1}$ — but that route gives $A=1, B=0$ too, confirming the answer. Check at $x=2$ : LHS $= \tfrac{9}{3} = 3$ ; RHS $= 2 + 1 = 3$ ✓.)

Example 4 — a mixed denominator (linear and irreducible quadratic)

Express $\dfrac{2x^2 + x + 3}{(x+1)(x^2+2)}$ in partial fractions.

The quadratic $x^2+2$ is irreducible (discriminant $-8<0$ ), so the form mixes a constant numerator over the linear factor and a linear numerator over the quadratic — three constants for the degree-3 denominator:

$\frac{2x^2 + x + 3}{(x+1)(x^2+2)} = \frac{A}{x+1} + \frac{Bx + C}{x^2+2}. \quad (\text{M1: form})$

Clear denominators:

$2x^2 + x + 3 = A(x^2+2) + (Bx + C)(x+1). \quad (\text{M1})$

Substitute the linear root $x=-1$ : $2 - 1 + 3 = 4 = A(1+2) = 3A$ , so $A = \tfrac43$ . (A1)

Compare $x^2$ : $2 = A + B$ , so $B = 2 - \tfrac43 = \tfrac23$ . (A1)

Compare constants: $3 = 2A + C$ , so $C = 3 - \tfrac83 = \tfrac13$ . (A1)

$\therefore\; \frac{2x^2 + x + 3}{(x+1)(x^2+2)} = \frac{4}{3(x+1)} + \frac{2x + 1}{3(x^2+2)}.$

(M1 form; M1 clear; A1 $A$ ; A1 $B$ ; A1 $C$ . Verify with the unused $x^1$ coefficient: RHS gives $B + C = \tfrac23 + \tfrac13 = 1$ , matching the LHS coefficient $1$ ✓. This example combines every form-selection rule in one fraction — the kind of synthesis a longer exam part demands.)

Specimen-style exam question

(specimen-style — not from any past paper)

(a) Express $\displaystyle f(x) = \frac{4}{(x+1)(x-1)^2}$ in partial fractions. (b) Hence find $\displaystyle \int f(x)\,dx$ .

(a) Degree-3 denominator → three constants:

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

$x = 1$ : $4 = C(2) \Rightarrow C = 2$ . $x = -1$ : $4 = A(-2)^2 = 4A \Rightarrow A = 1$ . Compare $x^2$ : $0 = A + B \Rightarrow B = -1$ .

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

Further Partial Fractions

Further Partial Fractions

Where this sits in AQA 7367

Core theory: matching the form to the denominator

Why a repeated factor needs every lower power

Why an irreducible quadratic gets a linear numerator

Two routes to the constants

The Cover-Up Rule, made concrete

Worked examples (with mark scheme)

Example 1 — a repeated linear factor

Example 2 — an irreducible quadratic factor

Example 3 — an improper fraction (divide first)

Example 4 — a mixed denominator (linear and irreducible quadratic)

Specimen-style exam question

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