Denominator:Denominator:Linear and different factorsLinear and different factors

Worked exampleWorked example

Type 1

Check top line is lower degree

Find partial fractions for

Factorise bottom line

A ( 2x – 3 )

B ( x + 2 )

+Split into commonsense fractions.A and B are constants.Note: top line must be lower degree

5x – 11 ( 2x – 3 )( x + 2 )

5x – 11 2x2 + x - 6

5x – 11 2x2 +x - 6

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

Combine the fractions

=

=

=

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

= 5x – 11 2x2 +x - 6

=>

=> 5x – 11 = A( x + 2 ) + B( 2x – 3 )

1: Equate coefficients and solve a pair of simultaneous equations.

Now find the unknowns A and B. There are two ways to do this.

2: Substitute “clever” values of x into the identity i.e. values of x which make terms disappear

We will use the second method.

Identity

Identity: 5x – 11 = A( x + 2 ) + B( 2x – 3 )

Let x = -2 ( x + 2 = 0 )

-10 – 11 = -7B -21 = -7B B = 3

( 2x - 3 = 0 )Let x = 32

– 11 = A72

15 2

15 – 22 = 7A -7 = 7A A =-1

……… x (2)

Partial fractions are:

5x – 11 2x2 +x - 6

-1 ( 2x – 3 )

+= 3 ( x + 2 )

Denominator:Denominator:Linear factor and an Linear factor and an

irreducible quadratic factorirreducible quadratic factor

Type 2

Worked exampleWorked example

Check top line is lower degree

Find partial fractions for

Factorise denominator fully

A ( x + 1 )

Bx + C ( x2 + 4 )

+Now ready to split into commonsense fractions. A, B and C are constants.Note: top line must be lower degree.

3x – 2( x + 1 )( x2 + 4 )

3x – 2 ( x + 1 )( x2 + 4 )

A( x2 + 4 ) +( Bx + C )( x + 1 ) ( x + 1 )( x2 + 4 )

Combine the fractions

=

=

=

3x – 2 ( x + 1 )( x2 + 4 )

Identity: 3x – 2 = A( x2 + 4 ) +( Bx + C )( x + 1 )

Let x = -1 ( x + 1 = 0 )

-3 – 2 = A((-1)2 + 4) -5 = 5A A = -1

Let x = 0–2 = 4A + C-2 = -4 + C C = 2

Partial fractions are:

We have run out of clever values. Now choose simple values, not already used, and use the values of the constants already found

Let x = 1

1 = 5A + ( B + C )( 2 )1 = -5 + 2B + 42B = 2 B = 1

(but A = -1 and C = 2)

= 3x – 2 ( x + 1 )( x2 + 4 )

-1 ( x + 1 )

x + 2 ( x2 + 4 )

+

(but A = -1)

Denominator:Denominator:Repeated linear factorsRepeated linear factors

Type 3

Worked exampleWorked example

Find partial fractions for 7x2 -11x - 5( x + 2 )( x - 1 )2

The factor ( x – 1 ) is repeated.There are 3 factors on the bottom line so we must split it into 3 bits.The correct way to split is as follows

A ( x + 2 )

B ( x - 1 )

+= 7x2 -11x - 5( x + 2 )( x - 1 )2

+ C ( x - 1 )2

Identity: 7x2 -11x - 5 = A( x - 1 )2 + B( x - 1 )( x + 2 ) + C ( x + 2 )clever values

Let x = 1 ( x - 1 = 0 )7 – 11 - 5 = C( 3 )

-9 = 3C C = -3

Let x = -2 ( x + 2 = 0 )28 + 22 - 5 = A( -3 )2

45 = 9A A = 5

We have run out of clever values. So choose simple values, not already used, and use the values of the constants already found.

Let x = 0–5 = A(-1)2 +B(-1)(2) + C(2)-5 = A – 2B + 2C ( but A = 5 and C = -3 )

-5 = 5 – 2B - 62B = 4 B = 2

Identity: 7x2 -11x - 5 = A( x - 1 )2 + B( x - 1 )( x + 2 ) + C ( x + 2 )

Partial fractions are:

7x2 -11x - 5( x + 2 )( x - 1 )2

5 ( x + 2 )

2 ( x - 1 )

+= - 3 ( x - 1 )2

3x2 + 2x + 9( x - 3 )( x - 2 )3

4 factors, so split into 4 bits

A (x - 3 )

B ( x - 2 )

+= + C ( x - 2 )2

+ D ( x - 2 )3

5x + 7 x2( x + 1 )

x is repeated. 3 factors, so split into 3 bits

A x

+= + C ( x + 1 )

B x2

f(x)( x + a )n n factors, so split into n bits

In general

A1 (x + a )

+= + + A2 (x + a )2

A3 (x + a )3

An (x + a )n

. . . . . . +

Numerator same degree or Numerator same degree or higher than the denominatorhigher than the denominator

Worked exampleWorked example

Find partial fractions for x( x + 3 ) x2 + x - 12

Numerator is degree 2, denominator is degree 2.Same degree => DIVIDE OUT first

x2 + x - 12

x2 + 3x + 0

1

x2 + x - 12 2x + 12 REMAINDER

QUOTIENT

=>

x( x + 3 ) x2 + x - 12

= 1 + 2 x + 12 x2 + x - 12

NOW FIND PARTIAL FRACTIONS FOR THIS BIT

THE FINAL ANSWER IS

x( x + 3 ) x2 + x - 12

= 1 + 18 7( x – 3 ) - 4

7( x + 4 )

Let’s do some more examples

Express in partial fractions. 3x2 + 2x + 1( x + 1 )(x2 + 2x + 2)

We must first use the discriminant to verify that the quadratic factor (x2 + 2x + 2) is irreducible.

For x2 + 2x + 2 :

a = , b = , c =

1 2 2

b2 – 4ac =

22 – 4(1)(2)

-4b2 – 4ac < 0 => x2 + 2x + 2 is

irreducible 3x2 + 2x + 1( x + 1 )(x2 + 2x + 2)

=

=

Another example:

Express as the sum of a polynomialand partial fractions.

x3 - 3x (x2 - x - 2)