PartialFractionsExpansion

Partial Fractions

Adding fractions

To add fractions we rewrite the fractions with a common denominator then add the numerators.

Example Find the sum of

3

3 4

2 5

x+ +x−

The common denominator of 3

3 4

2 5

x+ and x− is 3

b

x+4

gb g

x−5

Rewrite each fraction with a denominator of

b

3x+4

gb g

x−5

3

3 4

3 5

3 4 5

x

x x x

+ =

−

+ −

b g

b

gb g

and 2 5

2 3 4

3 4 5

x

x x x

− =

+

+ −

b

g

b

gb g

Add the numerators, then expand all brackets and simplify

Therefore

3 5

3 4 5

2 3 4

3 4 5

x x x

x x x

−

+ − +

+

+ −

b g

b

gb g

b

b

gb g

g

= − + +

+ −

3 5 2 3 4

3 4 5

x x

x x

b g b

g

b

gb g

= −

− −

9 7

3 2 11 20

x x x

3

3 4

2 5

x+ + x− =

−

− −

9 7

3 2 11 20

x x x

The reverse of this process is to split a fraction into partial fractions. In the above example

9 7 3 2 11 20

x x x

−

− − = + + −

3

3 4

2 5

x x

Algebraic fraction Partial fractions

The form of the numerator in the partial fractions depends only on the type of the factors in the denominator of the original fraction, as indicated below.

Each distinct linear factor eg. (x

−

a) has a

corresponding partial fraction of the form

(

)

A

x−a where A is a constant.

Each repeated linear factor eg. (x

−

a)2 has a

corresponding partial fractions of the form

(

) (

)

2

A B

x a x a

+

− − A

and B constants.

. Each quadratic factor eg. ax2+bx+chas a corresponding partial fraction of the form. 2

Ax B ax bx c

+

+ + A and B constants.

Examples The following fractions have been written in partial fraction form (without evaluating the constants in the numerators).

(1) 3 1

1 2 1 2

x

x x

A x

B x

+

− + = − + +

b gb

g b g b

g

linear factors only in the denominator

(2) 3 1

1 32 1 3 32

x x x

A x

B x

C x

+

+ +

=

+ + + + ₊

b gb g b g b g b g

repeated linear factor in the denominator gives rise to a partial fraction for both

x

b g

−3 and x

b g

−3 2 (3) 3 1

2 3 2 3

2 2

x x x x

A x

Bx C

x x

+

− +

= + +

− +

d

i

d

i

quadratic factor in the denominator

In general the numerator of a partial fraction is a polynomial of degree one less than the factor in the denominator.

Note the special case of repeated factors, example (2) above.

Exercise 1 Write the following in partial fraction form, but do not calculate the numerical values for the constants in the numerator. If possible, factorise the quadratic factor first.

(a) x

x x

+

− +

6

2 3 4

b

gb

g

(b) 5

4 2 3 2

Linear factors

The next task is to find values for the constants in the numerator. The first example will contain linear factors in the denominator only.

Example

Write x

x x

−

+ −

5

2 3

2 as the sum of partial fractions.

factorise the denominator

write the general expression for the partial fractions

add the fractions on the right side

the equation holds for all values of x (except x = 1 and x =

−

3) so we can

equate the numerators

expand brackets and collect like terms

equate coefficients of powers of x on both sides

solve these simultaneous equations to give

substitute for A and B the partial fractions are

x

x x

x

x x

−

+ − =

−

− +

5

2 3

5

1 3

2

b gb g

x

x x

A x

B x

−

− + = − + +

5

1 3 1 3

b gb g b g b g

A x

B x

A x B x x x

− + + =

+ + −

− +

1 3

3 1

1 3

b g b g

b g b g

b gb g

x x x

A x B x x x

−

− + =

+ + −

− +

5

1 3

3 1

1 3

b gb g

b g b g

b gb g

x− =5 A x

b g b g

+3 +B x−1

x− =5

b

A+B x

g b

+ 3A−B

g

A B A B

+ =

− = −

1

3 5

(coefficients of x) (constant terms)

A= −1 and B=2

− and

− +

1 1

2 3

x x

therefore x

x x

−

+ −

5

2 3

2 = ₊ − ₋

2 3

1 1

Example

Solving the simultaneous equations that result from equating coefficients can sometimes be quite lengthy. An alternative method is to equate the numerators and, before expanding the brackets, substitute a value of x into both sides of the equation so that only one variable remains. Repeat this to find other variables. This method will not necessarily find all variables, but will often make calculations easier.

Using the previous example, after equating numerators we had

x− =5 A x

b g b g

+3 +B x−1

substituting x = 1 will eliminate B and substituting x = −3 will eliminate A.

if x = 1 then if x = −3 then

1 − 5 = A(1 + 3) + B(1− 1) −3 −5 = A(−3 + 3) + B(−3 − 1) solving gives solving gives

A = − 1 B = 2

Repeated linear factors

In this example the denominator contains a repeated linear factor.

Example Express 5 3 1

3 2

2 3

x x x x

+ +

− − in terms of partial fractions.

factorise the denominator

write the general expression for the

partial fractions

add the fractions on the right side

equate numerators

5 3 1

3 2

5 3 1

1 2

2 3

2 2

x x x x

x x x x

+ +

− − =

+ +

+ −

b g b

g

5 3 1

3 2 2 1 ₁

2

3 2

x x x x

A x

B x

C x

+ +

− − =

b

−

g b g b g

+ + + ₊

A x

B x

C x

A x B x x C x

x x

− + + + ₊ =

+ + + − + −

− +

2 1 ₁

1 1 2 2

2 1

2

2

2

b

g b g b g

b g

b

b gb

gb g

g b

g

∴ 5 3 1

3 2

2 3

x x x x

+ +

− − =

A x B x x C x

x x

+ + + − + −

− +

1 1 2 2

2 1

2

2

b g

b gb

g b

g

b

gb g

to find A substitute x = 2

in the equation (the terms involving B and

C will equal zero and we can solve for A)

to find C substitute x =

−

1

( the terms involving A and B will equal zero)

to find B substitute the values already found (A=1 and C=

−

1) and a value for x (substituting x=0 keeps the arithmetic simple)

therefore

substitute for A, B, and C the partial fractions are

the solution is

5x2+3x+1 = A x

b g

+12 +B x

b gb

+1 x−2

g b

+C x−2

g

5 2 3 2 1 2 1 2 1 2 2 2 2

3 3 0 0

2 2

2

× + × + = + + + − + −

= + +

A B C

A B C

b g

b gb g b g

b g

b gb g b g

27 9

3 = =

A A

5x2+3x+1 = A x

b g

+12 +B x

b gb

+1 x−2

g b

+C x−2

g

5× −

b g

12+ × −3

b g

1 + =1 A

b g

0 2+B

b gb g b g

0 −3 +C −3 3 3

1 = −

= −

C C

5x2+3x+1 = A x

b g

+12 +B x

b gb

+1 x−2

g b

+C x−2

g

0 0 1+ + =3 1

b g

2+B

b gb g b g

1 −2 −1 2− 1 2 5

2

= − +

=

B B

A = 3, B =2, and C = − 1

(

)

(

)

(

)

2

3 2 1

, and

2 1 ₁

x− x+ _x+

5 3 1

3 2

3 2

2 1

1

1 2

3 2

x x

x x x x x

+ +

Quadratic or higher factor

The numerator for a quadratic factor has the form Ax+B. In general if the

denominator is of degree n then the numerator of the partial fraction is a polynomial of degree n

−

1.

Example Express

5 6 5

2 3 2

2

3 2

x x x x x

− +

− + − in terms of partial fractions.

factorise the denominator

write the general expression for the partial fractions

add the fractions on the right side

equate numerators

expand the brackets on the right-hand side of the equation

equate coefficients of powers of x

solve the simultaneous equations to give

substitute for A, B and C

5 6 5

2 3 2

5 6 5

1 2

2

3 2

2 2

x x

x x x

x x

x x x

− +

− + − =

− +

− − +

b gd

i

5 6 5

2 3 2 1 2

2

3 2 2

x x

x x x

A x

Bx C x x

− +

− + − = − +

+

− +

b g d

i

A x

Bx C x x

A x x Bx C x x x x

− +

+

− +

=

− + + + −

− − +

1 2

2 1

1 2

2

2

2

b g d

i

d

b gd

i b

i

gb g

5 6 5

2 3 2

2 1

1 2

2

3 2

2

2

x x x x x

A x x Bx C x x x x

− +

− + − =

− + + + −

− − +

d

i b

gb g

b gd

i

5x2−6x+5=A x

d

2−x+2

i b

+ Bx+C x

gb g

−1

(

)

(

)

(

)

2 2 2

2

5 6 5 2

2

x x Ax Ax A Bx Bx Cx C A B x C A B x A C

− + = − + + − + −

= + + − − + −

5

6

2 5

A B C A B

A C

+ =

− − = −

− =

A = 2, B = 3 , and C =

−

1

2

5x −6x+5

Exercise 2 Express the following as partial fractions.

(a) 2

3 2

x x− x+

b gb

g

(b) x

x x

+

− +

2

5 6

2

(c) x x

x x x

2 2

2 4

1 3 3

+ +

+ + +

b gd

i

(d)

3

1 2

x x

−

b g

(e) x x

x x x

2

3 2

2 2

− +

Answers

Exercise 1 (a)

2 3 4

A B

x− + x+ (b) 4 1 2

A B C D

x + x+ + x− +x−

(c) ₂

1 3

A Bx C x x

+ +

+ + (d)

(

2

) (

₂

)

2 2 3 7

A B Cx D

x _x x x

+

+ +

+ ₊ − +

Exercise 2 (a)

(

)

(

)

6 4

5 x−3 +5 x+2 (b)

5 4

3 2

x− −x− (c)

(

)

2

3 2 5

1 3 3

x

x x x

+ −

+ + +

(d)

(

)

2

(

)

3 3

1

1_−x − −x (e) 2

2 4

1

x x x x

+ −