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A2-Level Maths: Core 4for Edexcel

C4.1 Algebra and functions

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Partial fractions

Denominators with distinct linear factors

Denominators with a repeated linear factor

Improper fractionsCo

nte

nts

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Partial fractions

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Partial fractions

We know that two or more algebraic fractions can be added or subtracted to give a single fraction. For example:

Now, suppose we want to reverse the process.

2 1=

5 + 3x x

2( + 3) ( 5)

( 5)( + 3)

x x

x x

2 + 6 + 5=

( 5)( + 3)

x x

x x

+11=

( 5)( + 3)

x

x x

In other words, suppose we are given and

asked to express it as a sum of two separate fractions.

+11

( 5)( + 3)

x

x x

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Two distinct linear factors

For example:

This process is called expressing in partial fractions.

+11

( 5)( + 3)

x

x x

Express in partial fractions.7 1

( +1)( 3)

x

x x

The first step is to set up an identity.

The denominator of this fraction has two distinct linear factors so let

7 1+

( +1)( 3) +1 3

x A B

x x x x

where A and B are constants to be found.

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Two distinct linear factors

There are now two ways to continue:

using suitable substitutions,

by equating coefficients.

7 1+

( +1)( 3) +1 3

x A B

x x x x

7 1 ( 3)+ ( +1)

( +1)( 3) ( +1)( 3)

x A x B x

x x x x

To solve by substitution, we choose values of x that make the brackets zero.

7 1 ( 3)+ ( +1)x A x B x 1

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Two distinct linear factors

Therefore

7 1= ( 4)+ (0)A B

4 = 8A

= 2A

21 1= (0)+ (4)A B

4 = 20B

= 5B

7 1 2 5+

( +1)( 3) +1 3

x

x x x x

Here substitute x = –1 into : 1

Now substitute x = 3 into :1

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Partial fractions

Denominators with distinct linear factors

Denominators with a repeated linear factor

Improper fractionsCo

nte

nts

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Denominators with distinct linear factors

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Two distinct linear factors

Express in partial fractions.2

(3 2)(2 1)

x

x x

This time if we use the substitution method we’ll have to substitute fractional values for x.

Let’s multiply out the brackets and equate coefficients instead.

Let2

+(3 2)(2 1) 3 2 2 1

x A B

x x x x

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

This can be simplified by multiplying through by (3x – 2)(2x –1):

2 2 + 3 2x Ax A Bx B 2 (2 + 3 ) 2x A B x A B

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Two distinct linear factors

Now equate the constants:

Equate the coefficients of x:2 = 2 + 3A B 1

0 = 2A B 2

+ (2 × ) gives:21

2 = B= 2B

Substitute this into to find A: 1

2 = 2 6A 2 = 8A

= 4ATherefore 2 4 2

(3 2)(2 1) 3 2 2 1

x

x x x x

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Three distinct linear factors

Express as a sum of partial fractions.9 +1

( 3)( +1)(2 +1)

x

x x x

Let9 +1

+ +( 3)( +1)(2 +1) 3 +1 2 +1

x A B C

x x x x x x

Multiply through by (x – 3)(x +1)(2x +1):

This time we have three distinct linear factors, so:

27 +1= (4)(7)A

28 = 28A

=1A

To find A, substitute x = 3 into :1

9 +1 ( +1)(2 +1)+ ( 3)(2 +1)+ ( 3)( +1)x A x x B x x C x x 1

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Three distinct linear factors

9 +1= ( 4)( 1)B

8 = 4B

= 2B

We can now find C either by substituting x = or by reverting to the method of equating coefficients.

12

We can find B by substituting x = –1 into :1

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Three distinct linear factors

Therefore9 +1 1 2 2

+( 3)( +1)(2 +1) 3 +1 2 +1

x

x x x x x x

1= 3 3A B C

1=1+ 6 3C

3 = 6C

= 2C

We could also equate the coefficients of

x2 or x.

To avoid awkward arithmetic involving fractions, let’s form an equation in C by equating the constant terms in .1

9 +1 ( +1)(2 +1)+ ( 3)(2 +1)+ ( 3)( +1)x A x x B x x C x x 1

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Partial fractions

Denominators with distinct linear factors

Denominators with a repeated linear factor

Improper fractionsCo

nte

nts

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Denominators with a repeated linear factor

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Denominators with a repeated linear factor

In this case, the partial fractions will be of the form:

This is an example of a fraction whose denominator contains a repeated linear factor.

2

2 2

6 2+ +

( + 4)( 3) + 4 3 ( 3)

x x A B C

x x x x x

Suppose we wish to express in partial fractions.2

2

6 2

( + 4)( 3)

x x

x x

We can now find A, B and C using a combination of substitution and equating the coefficients.

2 2

2 2

6 2 ( 3) ( 4)( 3) ( 4)

( + 4)( 3) ( + 4)( 3)

x x A x B x x C x

x x x x

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Denominators with a repeated linear factor

2 26( 4) ( 4) 2 = ( 7)A 98 = 49A

= 2A

6(9) 3 2 = (7)C 49 = 7C

= 7C

To find B we can switch to the method of comparing coefficients.

2 26 2 ( 3) + ( + 4)( 3)+ ( + 4)x x A x B x x C x 1

Substitute x = 3 into :1

Substitute x = –4 into :1

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Denominators with a repeated linear factor

But A = 2 so:

6 = +A B

= 4B

Therefore2

2 2

6 2 2 4 7+ +

( + 4)( 3) + 4 3 ( 3)

x x

x x x x x

Express as a sum of partial fractions.2

13 +12

(4 )

x

x x

Let 2 2

13 +12+ +

(4 ) 4

x A B C

x x x x x

Equate the coefficients of x2 in :1

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Denominators with a repeated linear factor

Multiply through by x2(4 – x):

12 = 4B

= 3B

64 =16C

= 4C

Substitute x = 0 into :1

Substitute x = 4 into :1

213 +12 (4 )+ (4 )+x Ax x B x Cx 1

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Denominators with a repeated linear factor

We can find A by comparing the coefficients of x2.

0 = +A C=A C

But C = 4 so:

= 4A

Therefore 2 2

13 +12 4 3 4+ +

(4 ) 4

x

x x x x x

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Partial fractions

Denominators with distinct linear factors

Denominators with a repeated linear factor

Improper fractionsCo

nte

nts

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Improper fractions

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Improper fractions

Remember, an algebraic fraction is called an improper fraction if the degree of the polynomial is equal to, or greater than, the degree of the denominator.

To express an improper fraction in partial fractions we start by expressing it in the algebraic equivalent of mixed number form.

Any proper fractions contained in this form can then be expressed in partial fractions.

Express in partial fractions.2

2

2 3 +13

2 15

x x

x x

We can either use long division to divide 2x2 – 3x + 13 by x2 – 2x – 15 or we can set up an identity as follows:

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Improper fractions

Multiply through by (x + 3)(x – 5):

2

2

2 3 +13

2 15

x x

x x

22 3 +13

( + 3)( 5)

x x

x x

The numerator and the denominator are both of degree 2 and so they will divide to give a constant, A.

The part that is a proper fraction will have two distinct linear factors. So we can let

22 3 +13

( + 3)( 5)

x x

x x

+ +

( + 3) ( 5)

B CA

x x

Start by factorizing the denominator.

22 3 +13 ( + 3)( 5)+ ( 5)+ ( + 3)x x A x x B x C x 1

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Improper fractions

18 + 9 +13 = 8B8 = 40B

50 15 +13 = 8C8 = 48C

= 5B

= 6C

We can find A by equating the coefficients of x2 in . 1

Substitute x = –3 into :1

Substitute x = 5 into :1

= 2A

Therefore2

2

2 3 +13 5 62 +

2 15 ( + 3) ( 5)

x x

x x x x