partial fractions 43: partial fractions © christine crisp

Partial Fractions

43: Partial 43: Partial FractionsFractions

© Christine Crisp

Partial Fractions

)1)(2(

4233

xx

xx

We’ll start by adding 2 fractions. Cross and Smile

1

2

2

3

xxe.g.

)1)(2( xx

)1(3 x )2(2 x

)1)(2(

15

xx

x

The partial fractions for are)1)(2(

15

xx

x

1

2

2

3

xx

Partial Fractions

We have to find the values of A and B.

5 1

2 1( )( )

x

x x

12

x

B

x

A

To find the partial fractions, we start with

Partial Fractions

2x

A )1)(2( xx

1x

B )1)(2( xx

)1)(2(

15

xx

x )1)(2( xx

)1)(2(

15

xx

x12

x

B

x

A

Multiply by the denominator of the l.h.s.

)1(xA 15xSo,

If we understand the cancelling, we can in future go straight to this line from the 1st line.

To find the partial fractions, we start with

)2( xB

Partial Fractions

)2()1(15 xBxAx

The expressions are equal for all values of x, so I can choose to let x = 2.

Why should I choose x = 2 ?

ANS: x = 2 means the coefficient of B is zero, so B disappears and we can solve for A.

Partial Fractions

A 3

)2()1(15 xBxAx

2x

What value would you substitute next ?

ANS: Any value would do but x = 1 is good.

)22()12(1)2(5 BAA39

The expressions are equal for all values of x, so I can choose to let x = 2.

Partial Fractions

)2()1(15 xBxAx

2x )22()12(1)2(5 BAA39

1x )21()11(1)1(5 BAB36

)1)(2(

15

xx

xSo,

The expressions are equal for all values of x, so I can choose to let x = 2.

If we chose x = 1 instead, we get 4 = 2A – B, giving the same result.

A 3

B 2

12

xxA B

Partial Fractions

)2()1(15 xBxAx

2x )22()12(1)2(5 BAA39

1x )21()11(1)1(5 BAB36

)1)(2(

15

xx

xSo,

The expressions are equal for all values of x, so I can choose to let x = 2.

If we chose x = 1 instead, we get 4 = 2A – B, giving the same result.

12

xxAA B

A 3

B 2

Partial Fractions

)2()1(15 xBxAx

2x )22()12(1)2(5 BAA39

1x )21()11(1)1(5 BAB36

)1)(2(

15

xx

xSo,

The expressions are equal for all values of x, so I can choose to let x = 2.

If we chose x = 1 instead, we get 4 = 2A – B, giving the same result.

312

xx

2

A 3

B 2

Partial Fractions

)1)(3(

1

xxSolution: Let

13

x

B

x

A

Multiply by : )1)(3( xx

1 )1(xA

3x )2(1 A

1x )2(1 B21 A

21 B

)3( xBIt’s very important to write this each

time

e.g. 2 Express the following as 2 partial fractions.

)1)(3(

1

xx

Partial Fractions

We never leave fractions piled up like this, so

• The “halves” are written in the denominators ( as 2s ) and

• the minus sign is moved to the front of the 2nd fraction.

13)1)(3(

1 21

21

xxxxSo,

)1)(3(

1

xx )1(2

1

)3(2

1

xxFinally, we need to check the

answer.A thorough check would be to reverse the process and put the fractions together over a common denominator.

Partial Fractions

The method we’ve used finds partial fractions for expressions I’ll call Type 1

where, the denominator has 2 linear factors,

2)12)(3( x x

e.g.x

Partial Fractions

where, • the denominator has 2 linear factors,

The method we’ve used finds partial fractions for expressions I’ll call Type 1

2)12)(3( x x

e.g.x

( we may have to factorise to find them )

Partial Fractions

• and the numerator is a polynomial of lower degree than the denominator

The degree of a polynomial is given by the highest power of x.

The method we’ve used finds partial fractions for expressions I’ll call Type 1

2)12)(3( x x

e.g.x

where, • the denominator has 2 linear factors,

Partial Fractions

The degree of a polynomial is given by the highest power of x.

Here the numerator is of degree

The method we’ve used finds partial fractions for expressions I’ll call Type 1

2)12)(3( x x

e.g.x

1and the denominator of degree

• and the numerator is a polynomial of lower degree than the denominator

where, • the denominator has 2 linear factors,

Partial Fractions

where, the denominator has 2 linear factors,

The degree of a polynomial is given by the highest power of x.

and the numerator is a polynomial of lower degree then the denominator

The method we’ve used finds partial fractions for expressions I’ll call Type 1

2)12)(3( x x

e.g.x

and the denominator of degree

2Here the numerator is of degree

1

Partial Fractions

where, the denominator has 2 linear factors,

The degree of a polynomial is given by the highest power of x.

and the numerator is a polynomial of lower degree then the denominator

The method we’ve used finds partial fractions for expressions I’ll call Type 1

2)12)(3( x x

e.g.x

and the denominator of degree

2Here the numerator is of degree

1

Partial FractionsSUMMARYTo find partial fractions for expressions

like)12)(3(

2

xx

x

Let

123)12)(3(

2

x

B

x

A

xx

x

• Multiply by the denominator of the l.h.s.• Substitute a value of x that makes the coefficient of B equal to zero and solve for A.• Substitute a value of x that makes the coefficient of A equal to zero and solve for B.• Check the result by reversing the method or using the “cover-up” method.

Partial Fractions

Express each of the following in partial fractions.

1.

Exercises

)2)(3(

55

xx

x 2.)3)(12(

5

xx

3.

1

22 x

4.)1(

37

xx

x

Partial FractionsSolutions:1.

23)2)(3(

55

x

B

x

A

xx

x

)2)(3( xxMultiply by :)3()2(55 xBxAx

:2x B55 1 B:3x A520 4 A

2

1

3

4

)2)(3(

55

xxxx

xSo,

Partial FractionsSolutions:

)3)(12( xxMultiply by )12()3(5 xBxA

:3x B55 1 B:2

1x A255 2 A

3

1

12

2

)3)(12(

5

xxxxSo,

2.

312)3)(12(

5

x

B

x

A

xx

( I won’t write out any more checks but it is important to do them. )

Partial FractionsSolutions:

)1)(1( xxMultiply by )1()1(2 xBxA

:1x B22 1 B:1x A22 1 A

1

1

1

1

1

22

xxx

So,

3.11)1)(1(

2

1

22

x

B

x

A

xxx

Partial FractionsSolutions:

)1( xxMultiply by BxxAx )1(37

:0x A3:1x B 4 4 B

4.1)1(

37

x

B

x

A

xx

x

So,1

43

)1(

37

xxxx

x

Partial Fractions

If the denominator has 3 factors, we just extend the method.

e.g.

)2)(3)(1(

562

xxx

xx231

x

C

x

B

x

A

Solution: Multiply by )2)(3)(1( xxx

)3)(1()2)(1()2)(3(562 xxCxxBxxAxx

:1x )3)(4(561 A 1 AA1212 :3x )1)(4(5189 B 1 BB44

:2x )1)(3(5124 C 1 CC33

2

1

3

1

1

1

)2)(3)(1(

562

xxxxxx

xxSo,

Partial Fractions

The next type of fraction we will consider has a repeated linear factor in the denominator.e.g. 1

)2()1(

5342

2

xx

xx

We would expect the partial fractions to be

2)1()2()1(

53422

2

x

B

x

A

xx

xxor

either211)2()1(

5342

2

x

C

x

B

x

A

xx

xx

This is wrong because the first 2 fractions just give

1

x

BA, which is the same as having only one constant.

We will try this to see why it is also wrong.

)1(

)2(

Partial Fractions

2)1()2()1(

53422

2

x

B

x

A

xx

xxSuppose

Multiply by :)2()1( 2 xx22 )1()2(534 xBxAxx

:2x 2)3(5616 B 3 B

:1x A36 2 A

However,

BAx 250

Substituting B = 3 gives A = 1, an inconsistent result

We need 3 constants if the degree of the denominator is 3.

Partial Fractions

)2()1(

5342

2

xx

xxSo, for we need

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

53422

2

x

C

x

B

x

A

xx

xx

Partial Fractions

2)1()1(

5342

2

x

C

x

B

x

AxxUsing

Multiply by :)2()1( 2 xx

)2()1( 2 xx

Partial Fractions

2)1()1(

5342

2

x

C

x

B

x

AxxUsing

Multiply by :)2()1( 2 xx

)2()1( 2 xx

Partial Fractions

)2()1( 2 xx 2)1()1(

5342

2

x

C

x

B

x

AxxUsing

Multiply by :)2()1( 2 xx22 )1()2)(1()2(534 xCxxBxAxx

:2x 2)3(5616 C 3 C

:1x A3534 2 A

There is no other obvious value of x to use so we can choose any value.

e.g. :0x CBA 225Subst. for A and C:

3245 B 1 B

Partial Fractions

There is however, a neater way of finding B.

Since this is an identity, the terms on each side must be the same.

For example, we have on the l.h.s. so there must be on the r.h.s.

24x24x

We had 2x )1()2)(1()2(534 CBxAx x x x 2

Partial Fractions

There is however, a neater way of finding B.

Since this is an identity, the terms on each side must be the same.

2xSo, equating the coefficients of :

CB 4

For example, we have on the l.h.s. so there must be on the r.h.s.

24x24x

We had 2x )1()2)(1()2(534 CBxAx x x x 2

Partial Fractions

There is however, a neater way of finding B.

Since this is an identity, the terms on each side must be the same.

2xSo, equating the coefficients of :

CB 4

For example, we have on the l.h.s. so there must be on the r.h.s.

24x24x

We had 2x )1()2)(1()2(534 CBxAx x x x 2

Since 1,3 BCWe could also equate the coefficients of x ( but these are harder to pick out ) or the constant terms ( equivalent to putting x = 0 ).

Partial Fractions

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

53422

2

xxxxx

xxSo,

A B C

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

53422

2

xxxxx

xx 2 1 3

Partial Fractions

SUMMARY

• Let

21)1()2()1(

53422

2

x

C

x

B

x

A

xx

xx

To find partial fractions for expressions with repeated factors, e.g.

)2()1(

5342

2

xx

xx

• Check the answer by using a common denominator for the right-hand side.

• Work in the same way as for type 1 fractions, using the two obvious values of x and either any other value or the coefficients of .

2x

N.B. B can sometimes be zero.

Partial Fractions

ExercisesExpress each of the following in partial fractions.

1.

)1()2(

7102

2

xx

xx 2.

)12(

212

2

xx

xx

Partial FractionsSolutions:

Multiply by :)1()2( 2 xx22 )2()1)(2()1(710 xCxxBxAxx

1x 2)3(18 C 2 C

2x )3(7)2(10)2( 2 A

A39 3 ACoefficient of : 2x CB 1 1 B

1

2

2

1

)2(

3

)1()2(

71022

2

xxxxx

xxSo,

1. Let12)2()1()2(

71022

2

x

C

x

B

x

A

xx

xx

Partial FractionsSolutions:

Multiply by :)12(2 xx22 )12()12(21 CxxBxxAxx

0x A1

21x 2

212

21

21 )()()(21 C C4

147

Coefficient of : 2x CB 21 4 B

12

741

)1(

2122

2

xxxxx

xxSo,

2. Let12)12(

2122

2

x

C

x

B

x

A

xx

xx

7 C

Partial Fractions

If you are taking the OCR/MEI spec you can skip the next section.

SKIP

Partial Fractions

You may meet a question that combines algebraic division and partial fractions.

e.g. Find partial fractions for 23

1822

2

xx

xx

The degree of the denominator is equal to the degree of the numerator. Both are degree 2.

This is called an improper fraction.If the degree of the numerator is higher than the denominator the fraction is also improper.In an exam you are likely to be given the form of the partial fractions.

Partial Fractions

e.g. 1 Find the values of A, B and C such that

x

C

x

BA

xx

xx

112)1)(12(

652 2

Solution:

We don’t need to change our method

Multiply by :)1)(12( xx )12()1()1)(12(652 2 xCxBxxAxx

1x )3(652 C 3 C

21x )(6 2

325

21 B 2 B

:2x of Coef. A22 1 A

xxxx

xx

1

3

12

21

)1)(12(

652 2

So,

Partial Fractions

e.g. 2 Find the values of A, B and C such that

)3)(5()3)(5(

2 2

xx

CBxA

xx

x

Solution:Multiply by :)3)(5( xx

CBxxxAx )3)(5(2 2

CBx )5()5(25 2 CB 550

CBx )3()3(23 2 CB 318

CAx 1500 2 A

)1(

)2()2()1( B832 4 B

30 C)2(

So,)3)(5(

4302

)3)(5(

2 2

xx

x

xx

xIf you notice at the start, by looking at the terms on the l.h.s., that A = 2, the solution will be shorter as you can start with x = 0 and find C, then B.

2x

Partial Fractions

You will need to divide out but you will probably only need one stage of division so it will be easy.

If you aren’t given the form of the partial fractions, you just need to watch out for an improper fraction.

Partial Fractions

2323

18222

2

xxxx

xx )23( 2 xx2 x2

x8 x6 x2

e.g. 2

Partial Fractions

2323

18222

2

xxxx

xx )23( 2 xx2 x2 3

1 4 3

e.g. 2

Partial Fractions

2323

18222

2

xxxx

xx

So,

23

322

23

18222

2

xx

x

xx

xx

We can now find partial fractions for 23

322

xx

x

We get 2

7

1

5

23

322

xxxx

x

So,2

7

1

52

23

1822

2

xxxx

xx

e.g. 2

)23( 2 xx2 x2 3

Partial Fractions

22 22

2

xxxx

x 2 x

1. Express the following in partial fractions:

Exercise

22

2

xx

x

Solution: )2( 2 xx

2

21

2 22

2

xx

x

xx

x

Dividing out:

Partial Fractions

Partial Fractions:

21)2)(1(

2

x

B

x

A

xx

xLet

2

21

2 22

2

xx

x

xx

x

Multiply by :)2)(1( xx

)1()2(2 xBxAx

:2x )3(4 B 34 B

:1x )3(1 A 31 A

)2(3

4

)1(3

11

22

2

xxxx

xSo,

Partial Fractions

The next ( final ) section is only for those of you doing the OCR/MEI spec.

Partial Fractions

The 3rd type of partial fractions has a quadratic factor in the denominator that will not factorise.

e.g.

)1)(4(

52

2

xx

xx

The partial fractions are of the form

142

x

C

x

BAx

The method is no different but the easiest way to find A, B and C is to use the obvious value of x but then equate coefficients of the term and equate constants.

2x

Partial Fractions

1. Express the following in partial fractions:

Exercise

)1)(4(

52

2

xx

xx

Solution: 14)1)(4(

522

2

x

C

x

BAx

xx

xx

Multiply by :)1)(4( 2 xx

)4()1)((5 22 xCxBAxxx

:1x )5(5 C 1 C

Coefficient of :2x CA 1 2 AConstants: CB 45 1 B

Partial Fractions

1

1

4

12

)1)(4(

522

2

xx

x

xx

xxSo,

Partial Fractions

Partial Fractions

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

Partial Fractionse.g. Express the following as 2 partial fractions.

)1)(3(

1

xx

)1)(3(

1

xxSolution: Let

13

x

B

x

A

Multiply by : )1)(3( xx

1 )1(xA

3x )2(1 A

1x )2(1 B21 A

21 B

)3( xB

It’s very important to write this each time

Partial Fractions

We never leave fractions piled up like this, so

• The “halves” are written in the denominators ( as 2s ) and

• the minus sign is moved to the front of the 2nd fraction.

13)1)(3(

1 21

21

xxxxSo,

)1)(3(

1

xx )1(2

1

)3(2

1

xxFinally, we need to check the

answer.A thorough check would be to reverse the process and put the fractions together over a common denominator.

Partial Fractions

)1)(3(

1

xx )1()3(

x

B

x

AAnother check is to use the “cover-up” method:

We get

)13)(3(

1

xTo check B, substitute x = in the l.h.s. but cover-up)1( x

2

1

Cover-up on the l.h.s. and substitute x = 3 into the l.h.s. only

)3( x

2

1

)( A

)( B )1)(31(

1

x

To check A, find the value of x that makes the factor under A equal to zero( x = 3 )

Partial Fractions

SUMMARYTo find partial fractions for expressions

like)12)(3(

2

xx

x

Let

123)12)(3(

2

x

B

x

A

xx

x

• Multiply by the denominator of the l.h.s.• Substitute a value of x that makes the coefficient of B equal to zero and solve for A.• Substitute a value of x that makes the coefficient of A equal to zero and solve for B.• Check the result by reversing the method or using the “cover-up” method.

Partial Fractions

The next type of fraction we will consider has a repeated linear factor in the denominator.e.g. 1

)2()1(

6332

2

xx

xx

We would expect the partial fractions to be

2)1()2()1(

63322

2

x

B

x

A

xx

xxor

either211)2()1(

6332

2

x

C

x

B

x

A

xx

xx

This is wrong because the first 2 fractions just give

1

x

BA, which is the same as having only one constant.

We will try this to see why it is also wrong.

)1(

)2(

Partial Fractions

2)1()2()1(

53422

2

x

B

x

A

xx

xxSuppose

Multiply by :)2()1( 2 xx22 )1()2(534 xBxAxx

:2x 2)3(5616 B 3 B

:1x A36 2 A

However,

BAx 250

Substituting A = 2 gives B = 1, an inconsistent result

We need 3 constants if the degree of the denominator is 3.

Partial Fractions

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

53422

2

x

C

x

B

x

A

xx

xxUsing

Multiply by :)2()1( 2 xx22 )1()2)(1()2(534 xCxxBxAxx

:2x 2)3(5616 C 3 C

:1x A3534 2 A

There is no other obvious value of x to use so we can choose any value.

e.g. :0x CBA 225Subst. for A and C:

3245 B 1 B

Partial Fractions

There is however, a neater way of finding B.

Since this is an identity, the terms on each side must be the same.

2xSo, equating the coefficients of :

CB 4

Since 1,3 BC

For example, we have on the l.h.s. so there must be on the r.h.s.

24x24x

We had 2x )1()2)(1()2(534 CBxAx x x x 2

We could also equate the coefficients of x ( but these are harder to pick out ) or the constant terms.

Partial Fractions

SUMMARY

• Let

21)1()2()1(

53422

2

x

C

x

B

x

A

xx

xx

To find partial fractions for expressions with repeated factors, e.g.

)2()1(

5342

2

xx

xx

• Check the answer by using using a common denominator for the right-hand side.

• Work in the same way as for type 1 fractions, using the two obvious values of x and either any other value or the coefficients of .

2x

N.B. B can sometimes be zero.

Partial Fractions

You may meet a question that combines algebraic division and partial fractions.

e.g. Find partial fractions for 23

1822

2

xx

xx

The degree of the denominator is equal to the degree of the numerator. Both are degree 2.

This is called an improper fraction.If the degree of the numerator is higher than the denominator the fraction is also improper.In an exam you are likely to be given the form of the partial fractions.

Partial Fractions

e.g. 1 Find the values of A, B and C such that

x

C

x

BA

xx

xx

112)1)(12(

652 2

Solution:

We don’t need to change our method

Multiply by :)1)(12( xx )12)1()1)(12(652 2 xCxBxxAxx

1x )3(652 C 3 C

21x )(6 2

325

21 B 2 B

:2x of Coef. A22 1 A

xxxx

xx

1

3

12

21

)1)(12(

652 2

So,

Partial Fractions

You will need to divide out but you will probably only need one stage of division so it will be easy.

If you aren’t given the form of the partial fractions, you just need to watch out for an improper fraction.

Partial Fractions

2323

18222

2

xxxx

xx

So,

23

322

23

18222

2

xx

x

xx

xx

We can now find partial fractions for 23

322

xx

x

We get 2

7

1

5

23

322

xxxx

x

So,2

7

1

52

23

1822

2

xxxx

xx

e.g. 2

)23( 2 xx2 x2 3

Partial Fractions

OCR/MEI onlyThe 3rd type of partial fractions has a quadratic factor in the denominator that will not factorise.

e.g.

)1)(4(

52

2

xx

xx

The partial fractions are of the form

142

x

C

x

BAx

The method is no different but the easiest way to find A, B and C is to use the obvious value of x but then equate coefficients of the term and equate constants.

2x