partial fractions. in the presentation on algebraic fractions we saw how to add 2 algebraic...
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Partial FractionsPartial Fractions
Partial Fractions
In the presentation on algebraic fractions we saw how to add 2 algebraic fractions.
To find partial fractions for an expression, we need to reverse the process of adding fractions.
We will also develop a method for reducing a fraction to 3 partial fractions.
Partial Fractions
)1)(2(
4233
xx
xx
We’ll start by adding 2 fractions.
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
The expressions are equal for all values of x so we have an identity.
The identity will be important for finding the values of A and B.
)1)(2(
15
xx
x12
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
This is where the identity is important.
)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
This is where the identity is important.
)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
This is where the identity is important.
)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
This is where the identity is important.
)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
This is where the identity is important.
)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
)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
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,
5
20
Check:
gives3x
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,
5
20
5
5gives2x,4Check
:gives3x
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,
5
20
5
5gives2x,4 1Check
:gives3x
( you don’t need to write out the check in full )
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
It would also be correct to write
2)1()2()1(
53422
2
x
C
x
BAx
xx
xx
but the fractions are not then reduced to the simplest form
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 225
Subst. 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 BC
We 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
Partial Fractions
)2()1(
)2(22
xx
x
2)1()1()2()1(
53422
2
xxxxx
xxSo,
We get
The “cover-up” method can only be used to check A and C so for a proper check we need to put the r.h.s. back over a common denominator.
So the numerator gives: )12(3)2(42 22 xxxxx
534 2 xx
2)1(3 x )2)(1(1 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
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 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,