What is 2.10: Partial fractions?

Factor the denominator first and assign one partial-fraction term to each linear factor. A repeated factor needs a term for every power: (x-a)^2 requires A/(x-a)+B/(x-a)^2.

How many marks are 2.10 questions worth?

2.10 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 2.10 appeared in?

We have not yet indexed a verified past-paper appearance for 2.10 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

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).

Draft — not yet indexed

Partial fractions

Worked answers and methods for 2.10 on Edexcel A-level Maths 9MA0.

Explanation

Factor the denominator first and assign one partial-fraction term to each linear factor. A repeated factor needs a term for every power: $(x-a)^2$ requires $A/(x-a)+B/(x-a)^2$ .
After multiplying through by the common denominator, substitute convenient roots or compare coefficients to determine the constants.
If the numerator degree is at least the denominator degree, divide first; omitting a repeated-factor term produces an identity that cannot hold.
Recombine the fractions to check the identity.
The simpler terms can then be differentiated, integrated or expanded as binomial/geometric series within the required domain.

Worked example

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

1.Let $(5x+1)/((x-1)(x+2))=A/(x-1)+B/(x+2)$ .
2.Hence $5x+1=A(x+2)+B(x-1)$ .
3.Substituting $x=1$ gives $6=3A$ , so $A=2$ .
4.Substituting $x=-2$ gives $-9=-3B$ , so $B=3$ .

Answer: $\dfrac2{x-1}+\dfrac3{x+2}$

Common mistakes

Don't write only one term for a repeated factor $(x-a)^2$ instead of both $A/(x-a)$ and $B/(x-a)^2$ .
Don't use an unnecessary linear numerator over a linear factor instead of a constant numerator.

Exam tip

Factor the denominator first, write every required fraction term, then multiply through before substituting convenient values.

Worked practice

Tier 1 · Easy

Express

\dfrac7{(x+1)(x+2)}

as partial fractions.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\dfrac7{x+1}-\dfrac7{x+2}$	3
Notes
Write $7/((x+1)(x+2))=A/(x+1)+B/(x+2)$ . Then $7=A(x+2)+B(x+1)$ . Setting $x=-1$ gives $A=7$ , while $x=-2$ gives $B=-7$ .

(3 marks)

Tier 2 · Standard

Express

\dfrac{3x+5}{(x+1)^2}

in the form

\dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}

, where

A

and

B

are constants.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$\dfrac3{x+1}+\dfrac2{(x+1)^2}$	3
Notes
Multiplying by $(x+1)^2$ gives $3x+5=A(x+1)+B=Ax+(A+B)$ . Comparing coefficients gives $A=3$ and $A+B=5$ , so $B=2$ . Therefore the decomposition is $3/(x+1)+2/(x+1)^2$ .

(3 marks)

Tier 3 · Hard

Express

\dfrac{5x+1}{(x-1)^2(x+2)}

as three partial-fraction terms.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$\dfrac1{x-1}+\dfrac2{(x-1)^2}-\dfrac1{x+2}$	5
Notes
Set the expression equal to $A/(x-1)+B/(x-1)^2+C/(x+2)$ . Multiplying through gives $5x+1=A(x-1)(x+2)+B(x+2)+C(x-1)^2$ . With $x=1$ , $6=3B$ , so $B=2$ ; with $x=-2$ , $-9=9C$ , so $C=-1$ . Comparing the $x^2$ coefficients gives $A+C=0$ , hence $A=1$ .

(5 marks)

Tier 1 · Easy

Given that

\dfrac{7x-1}{(x-3)^2}\equiv\dfrac{A}{x-3}+\dfrac{B}{(x-3)^2}

, find

B

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$B=20$	2
Notes
Multiplying by $(x-3)^2$ gives $7x-1=A(x-3)+B$ . Substituting $x=3$ gives $B=21-1=20$ .

(2 marks)

Tier 2 · Standard

Express

\dfrac{2x^2+6x-2}{x(x-1)(x+2)}

as three partial-fraction terms.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$\dfrac1x+\dfrac2{x-1}-\dfrac1{x+2}$	4
Notes
Set the expression equal to $A/x+B/(x-1)+C/(x+2)$ . Multiplying through gives $2x^2+6x-2=A(x-1)(x+2)+Bx(x+2)+Cx(x-1)$ . Substituting $x=0,1,-2$ gives $A=1$ , $B=2$ and $C=-1$ respectively.

(4 marks)

Tier 3 · Hard

Decompose

\dfrac{2x^3-2x^2+x+3}{x(x-1)^2}

fully into partial fractions, carrying out algebraic division first.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$2+\dfrac3x-\dfrac1{x-1}+\dfrac4{(x-1)^2}$	6
Notes
Since $x(x-1)^2=x^3-2x^2+x$ , division gives quotient $2$ and remainder $2x^2-x+3$ . Write the proper fraction as $A/x+B/(x-1)+C/(x-1)^2$ . Then $2x^2-x+3=A(x-1)^2+Bx(x-1)+Cx$ . Setting $x=0$ gives $A=3$ , setting $x=1$ gives $C=4$ , and comparing $x^2$ coefficients gives $A+B=2$ , so $B=-1$ . Combining with the quotient gives the stated decomposition.

(6 marks)

Tier 2 · Standard

Express

\dfrac{8x+3}{(2x-1)(x+3)}

in partial fractions.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$\dfrac2{2x-1}+\dfrac3{x+3}$	4
Notes
Write the expression as $A/(2x-1)+B/(x+3)$ . Multiplying through gives $8x+3=A(x+3)+B(2x-1)$ . Setting $x=1/2$ gives $7=7A/2$ , so $A=2$ . Setting $x=-3$ gives $-21=-7B$ , so $B=3$ .

(4 marks)

Tier 3 · Hard

The identity

\dfrac{x^2+kx+10}{(x-1)(x+2)(x-3)}\equiv\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{x-3}

has

A=C

. Find

k

and hence determine

A

B

and

C

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$k=-8$ $A=-\dfrac12$ , $B=2$ , $C=-\dfrac12$	6
Notes
Multiplying through and substituting $x=1$ gives $A=-(k+11)/6$ . Substituting $x=3$ gives $C=(3k+19)/10$ . Since $A=C$ , $-5(k+11)=3(3k+19)$ , so $-14k=112$ and $k=-8$ . Then $A=C=-1/2$ . Substituting $x=-2$ gives $30=15B$ , so $B=2$ .

(6 marks)

Tier 3 · Hard

Express

\dfrac{10x^2-3x-4}{(2x-1)^2(x+1)}

in partial fractions.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\dfrac3{2x-1}-\dfrac2{(2x-1)^2}+\dfrac1{x+1}$	5
Notes
Write the expression as $A/(2x-1)+B/(2x-1)^2+C/(x+1)$ . Multiplying through gives $10x^2-3x-4=A(2x-1)(x+1)+B(x+1)+C(2x-1)^2$ . Setting $x=1/2$ gives $-3=(3/2)B$ , so $B=-2$ . Setting $x=-1$ gives $9=9C$ , so $C=1$ . Comparing $x^2$ coefficients gives $10=2A+4C$ , hence $A=3$ .

(5 marks)

Q10

Tier 3 · Hard

10.

The decomposition of

\dfrac{2x^2+kx+5}{(x-1)(x+2)^2}

has the form

\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}

. Given that

B=0

, find

k

A

and

C

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$k=11$ , $A=2$ and $C=3$	5
Notes
Multiplying through gives $2x^2+kx+5=A(x+2)^2+B(x-1)(x+2)+C(x-1)$ . Since $B=0$ , comparison with $A(x+2)^2+C(x-1)$ gives $A=2$ from the $x^2$ coefficient. The constant term then gives $5=4A-C$ , so $C=3$ . Comparing the $x$ coefficients gives $k=4A+C=11$ . Thus the decomposition is $2/(x-1)+3/(x+2)^2$ .

(5 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 2.10. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in 2 Algebra and functions

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