In this chapter you will learn how an algebraic fraction can be written as a sumof two (or more) simpler fractions called partial fractions. You will apply this toareas of calculus and algebra in later chapters.

You learned how to combine simpler fractions into more complex ones in the C3 book, andSection 1.1 reviews this. Splitting complex fractions into partial fractions is the oppositeprocess. Different methods are required depending on the kind of fraction you are startingfrom.

Partial fractions11

1

1.1 You can add (or subtract) two or more simple fractions as long as thedenominators are the same.

Example

Calculate: a 13

38

b x

23

(x

11)

1

a 31

83

284 2

94

8

2

49

2174

b (x

23)

(x

11)

(x

2(3x

)

(x1

)1)

(x

1(x3

)(x3

)1)

2(x

(x

1)3

)(x1(

x 1)

3)

2(xx

23

)(x1x

1)3

(x

x3

)(x1 1)

The lowest common multiple of 3 and 8 is 24.

Add the numerators.

Add the numerators.

Expand the brackets.

Simplify the numerator.

The lowest common multiple is (x 3)(x 1),so change both fractions so that thedenominators are (x 3)(x 1).

Multiply 13 top and bottom by 8 to express in24ths.Multiply 38 top and bottom by 3 to express in24ths.

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2CHAPTER 1

Exercise

Express each of the following as a single fraction:

13

14

34

25

3x

x

21

(x

21)

(x

32)

(2x

4 1)

(x 2

1)

(x 7

3)

(x 2

4)

23x

(x 6

1)

3x

(x

21)

(x

12)

34x

(x 2

2)

(2x1 1)

(x

31)

(x

21)

(x

43)

10

987

654

321

1A

1.2 You can split a fraction with two linear factors in the denominator into partialfractions.

For example, (x

x1

)(x1 3)

is (x

23)

(x

11)

when split into partial fractions (see

Example 1).

In general, an expression with two linear terms in the denominator such as (x 3

1)1(x 2)

can be split into partial fractions of the form (x

A3)

(x

B2)

, where A and B are constants.

There are two methods of achieving this: by substitution and by equating coefficients.

Example

Split (x

63

x)

(x2 1)

into partial fractions by a substitution b equating coefficients.

2

a (x

6x3)

(x2 1)

(x

A3)

(x

B1)

6x 2 A(x 1) B(x 3)6 3 2 A(3 1) B(3 3)

16 4AA 4

6 1 2 A(1 1) B(1 3)8 4B

B 2

(x

6x3)

(x2 1)

(x

43)

(x

21)

A(x 1) B(x 3)

(x 3)(x 1)

Add the two fractions.

To find B substitute x 1.

To find A substitute x 3.

Because this is an equivalence relation setthe numerators equal to each other.

Set(x

63

x)

(x2 1)

identical to

(x

A3)

(x

B1)

.

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3Partial fractions

Exercise

Put the following into partial fraction form:

a b

c d

e f

g h

Show that can be written in the form where A and B

are constants to be found.

B(2 x)

A(4 x)

2x 5(4 x)(2 x)

2

2x 14x2 2x 15

8 xx2 4x

7 3xx2 3x 4

6x 6x2 9

2x 13(2x 1)(x 3)

7x 122x(x 4)

2x 11(x 1)(x 4)

6x 2(x 2)(x 3)

1

1B

b (x

6x3)

(x2 1)

(x

A3)

(x

B1)

6x 2 A(x 1) B(x 3) Ax A Bx 3B (A B)x (A 3B)

Equate coefficients of x:6 A B 1

Equate constant terms:2 A 3B 2

1 2 :8 4B

B 2Substitute B 2 in 1 6 A 2

A 4

A(x 1) B(x 3)

(x 3)(x 1)

1.3 You can also split fractions that have more than two linear factors on thedenominator into partial fractions.

An expression with three or more linear terms in the denominator such as

can be split into , and so on if there

are more terms.

C(x 4)

B(x 3)

A(x 1)

4(x 1)(x 3)(x 4)

Solve simultaneously.

Set (x

63

x)

(x2 1)

identical to

(x

A3)

(x

B1)

.

Add the two fractions.

Because this is an equivalence relation setthe numerators equal to each other.

Expand the brackets.

Collect like terms.

You want (A B)x A 3B 6x 2.Hence coefficient of x is 6, and constantterm is 2.

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4CHAPTER 1

Example

Express in partial fractions.6x2 5x 2x(x 1)(2x 1)

3

Let x6(x

x

2

1)5(2

xx

21)

Ax

(x

B1)

(2x

C 1)

6x2 5x 2 A(x 1)(2x 1) Bx(2x 1) Cx(x 1)

Let x 16 5 2 0 B 1 3 0

9 3BB 3

Let x 00 0 2 A 1 1 0 0

2 1AA 2

Let x 21

64

52 2 0 0 C 2

1 32

3 34CC 4

So x6(x

x

2

1)5(2

xx

21)

2x

x

31

2x

4 1

A(x 1)(2x 1) Bx(2x 1) Cx(x 1)

x(x 1)(2x 1)

Exercise

Express the following in partial fraction form:

a b c

By firstly factorising the denominator, express the following in partial fraction form:

a b c5x2 15x 8x3 4x2 x 6

5x2 15x 8x3 3x2 2x

6x2 7x 3

x3 x

2

5x2 19x 32(x 1)(x 2)(x 5)

10x2 8x 2x(2x 1)(3x 2)

2x2 12x 26(x 1)(x 2)(x 5)

1

1C

The denominators must be x, (x 1)and (2x 1).

Add the fractions.

Set the numerators equal.

Proceed by substitution OR byequating coefficients.

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5Partial fractions

An expression with repeated linear terms such as can be split into the form

.

Example

Split into partial fraction form.11x2 14x 5(x 1)2(2x 1)

4

C(x 3)2

B(x 3)

A(x 1)

6x2 29x 29(x 1)(x 3)2

1.4 You can express a fraction that has repeated linear terms in its denominator inpartial fraction form.

Let (1x1x

2

1)21(42xx

51)

(x

A1)

(x

B1)2

(2x

C 1)

Hence 11x2 14x 5 A(x 1)(2x 1) B(2x 1) C(x 1)2

11 14 5 A 0 B 1 C 02 1BB 2

141 7 5 A 0 B 0 C 4

1

34 4

1 C

C 311 2A C11 2A 3

2A 8A 4

Hence (1x1x

2

1)21(42xx

51)

(x

41)

(x

21)2

(2x

3 1)

A(x 1)(2x 1) B(2x 1) C(x 1)2

(x 1)2(2x 1)

Exercise

Put the following into partial fraction form:

x2 5x 7

(x 2)38

2x(x 2)2

72x2 2x 18x3 6x2 9x

65x2 2x 1

x3 x25

7x2 42x 64

x(x 4)24

2x2 2x 18

x(x 3)23

x2 10x 5(x 1)2(x 1)

23x2 x 2

x2(x 1)1

1D

Substitute C 3.

You need denominators of (x 1), (x 1)2and (2x 1).

Add the three fractions.

Set the numerators equal.

To find B substitute x 1.

To find C substitute x 12.

Equate coefficients of x2 to find A.

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6CHAPTER 1

An algebraic fraction is improper when the degree of the numerator is equal to, or largerthan, the degree of the denominator. An improper fraction must be divided first to obtain amixed number fraction before it can be expressed in partial fractions.

x(x

x

2

3),

(

xx

3

14)

x(x

2

32)

and (x 1)

x2

4

(x 2) are all examples of improper fractions.

Example

Express (3xx

2

1)3(xx

22)

in partial fractions.

5

1.5 You can split improper fractions into partial fractions by dividing thenumerator by the denominator.

(3xx

2

1)(3xx

22)

3xx2

2

33xx

22

3 x2 3x 23x2 3x 2

3x2 9x 66x 8

Therefore

(3xx

2

1)(3xx

22)

3 x2

6

x3

x8 2

3 (x

6x1)

(x8 2)

Let (x

6x1)

(x8 2)

(x

A1)

(x

B2)

6x 8 A(x 2) B(x 1)12 8 A 0 B 1

B 46 8 A 1 B 0

A 2

(3xx

2

1)(3xx

22)

3 (x

6x1)

(x8 2)

3 (x

21)

(x

42)

A(x 2) B(x 1)

(x 1)(x 2)

Multiply out the denominator.

Divide the denominator into the numerator.

Substitute x 2 to find B.

Factorise x2 3x 2.

Set the numerators equal.

Substitute x 1 to find A.

Write out full solution.

It goes in 3 times, with a remainder of 6x 8.

Write (3xx

2

1)3(xx

22)

as a mixed number

fraction.

Add the two fractions.

The denominators must be (x 1) and (x 2).

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7Partial fractions

Exercise

Express the following improper fractions in partial fraction form:

a b

c d

By factorising the denominator, express the following in partial fraction form:

a b

Show that can be expressed in the form

A Bx , where A, B, C and D are constants to be found.

Mixed exercise

Express the following in partial fraction form:

a b

c d

Write the following algebraic fractions in partial fraction form:

a b

c d

Given that f(x) 2x3 9x2 10x 3:

a Show that 3 is a root of f(x).

b Express in partial fractions.

(adapted)

E10f(x)

3

x4x2 2x 1

3x2 12x 8

(x 2)3

2x2 2x 8x2 2x 3

3x 1x2 2x 1

2

x2 1x(x 2)

15x 21(x 2)(x 1)(x 5)

7x2 2x 2

x2(x 1)x 3x(x 1)

1

1F

D(x 3)

C(x 1)

3x3 4x2 19x 8

x2 2x 33

x4 4x3 9x2 17x 12

x3 4x2 4x4x2 17x 11

x2 3x 4

2

2x2 1(x 1)2

x3 x2 x 3

x(x 1)

x2 10(x 2)(x 1)

x2 3x 2(x 1)(x 3)

1

1E

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8CHAPTER 1

Summary of key points1 An algebraic fraction can be written as a sum of two or more simpler fractions. This

technique is called splitting into partial fractions.

2 An expression with two linear terms on the denominator such as (x 3

1)1(x 2) can be

split by converting into the form (x

A3)

(x

B2)

.

3 An expression with three or more linear terms such as can be

split by converting into the form (x

A1)

(x

B3)

(x

C4)

.

4 An expression with repeated terms on the denominator such as can

be split by converting into the form (x

A1)

(x

B3)

(x

C3)2.

5 An improper fraction is one where the index of the numerator is equal to or higherthan the index of the denominator. An improper fraction must be divided first toobtain a mixed number fraction before you can express it in partial fractions.

For example, xx

2

2

33

xx

42

1 x2 3

2x 2 1

(x A

1)

(x B

2).

6x2 29x 29(x 1)(x 3)2

4(x 1)(x 3)(x 4)

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