1. Denominator: Linear and different factors
Partial Fractions
Type 1
Denominator: Linear and different factors
Worked example

2. Find partial fractions for
5x – 11
2x2 + x - 6
Find partial fractions for
5x – 11
2x2 +x - 6
Check top line is lower degree
5x – 11
( 2x – 3 )( x + 2 )
Factorise bottom line =
Split into commonsense fractions.
A and B are constants.
Note: top line must be lower degree A
( 2x – 3 ) B
( x + 2 ) = +
A( x + 2 ) + B( 2x – 3 )
( 2x – 3 )( x + 2 )
Combine the fractions =

3. Now find the unknowns A and B.
5x – 11
2x2 +x - 6
A( x + 2 ) + B( 2x – 3 )
( 2x – 3 )( x + 2 )
=> =
=>
5x – 11 =
Identity
A( x + 2 ) + B( 2x – 3 )
Now find the unknowns A and B.
There are two ways to do this.
1: Equate coefficients and solve a pair of simultaneous
equations.
2: Substitute “clever” values of x into the identity
i.e. values of x which make terms disappear
We will use the second method.

4. Partial fractions are:
Identity:
5x – 11 =
A( x + 2 ) + B( 2x – 3 )
Let x = 3 2
Let x = -2
( x + 2 = 0 )
( 2x - 3 = 0 )
– 11 = A 7 2 15
-10 – 11 = -7B
-21 = -7B
B = 3
……… x (2)
15 – 22 = 7A
-7 = 7A
A =-1
Partial fractions are:
5x – 11
2x2 +x - 6 -1
( 2x – 3 ) 3
( x + 2 ) = +

5. Denominator: Linear factor and an irreducible quadratic factor
Partial Fractions
Type 2
Denominator: Linear factor and an irreducible quadratic factor
Worked example

6. Find partial fractions for
3x – 2
( x + 1 )( x2 + 4 )
Find partial fractions for
3x – 2
( x + 1 )( x2 + 4 )
Check top line is lower degree
3x – 2
( x + 1 )( x2 + 4 )
Factorise denominator fully =
Now ready to split into commonsense fractions. A, B and C are constants.
Note: top line must be lower degree. A
( x + 1 )
Bx + C
( x2 + 4 ) = +
A( x2 + 4 ) +( Bx + C )( x + 1 )
( x + 1 )( x2 + 4 )
Combine the fractions =

7. Partial fractions are:
Identity:
3x – 2 =
A( x2 + 4 ) +( Bx + C )( x + 1 )
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
( x + 1 = 0 )
-3 – 2 = A((-1)2 + 4)
-5 = 5A
A = -1
(but A = -1
and C = 2)
Let x = 1
Let x = 0
–2 = 4A + C
(but A = -1)
1 = 5A + ( B + C )( 2 )
-2 = -4 + C
C = 2
1 = B + 4
2B = 2
B = 1
Partial fractions are:
3x – 2
( x + 1 )( x2 + 4 ) -1
( x + 1 )
x + 2
( x2 + 4 ) = +

8. Denominator: Repeated linear factors
Partial Fractions
Type 3
Denominator: Repeated linear factors
Worked example

9. Find partial fractions for
7x2 -11x - 5
( x + 2 )( x - 1 )2
Find partial fractions for
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
7x2 -11x - 5
( x + 2 )( x - 1 )2 A
( x + 2 ) B
( x - 1 ) 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 )
Let x = -2
( x + 2 = 0 )
7 – = C( 3 )
= A( -3 )2
-9 = 3C
45 = 9A
C = -3
A = 5

10. Partial fractions are:
Identity:
7x2 -11x - 5 =
A( x - 1 )2 + B( x - 1 )( x + 2 ) + C ( x + 2 )
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 - 6
2B = 4
B = 2
Partial fractions are:
7x2 -11x - 5
( x + 2 )( x - 1 )2 5
( x + 2 ) 2
( x - 1 ) 3
( x - 1 )2 = + -

11. More splitting examples
3x2 + 2x + 9
( x - 3 )( x - 2 )3
4 factors, so split into 4 bits A
(x - 3 ) B
( x - 2 ) D
( x - 2 )3 = C
( x - 2 )2 + + +
5x + 7
x2( x + 1 )
x is repeated. 3 factors, so split into 3 bits A x B x2 C
( x + 1 ) = + +
In general
f(x)
( x + a )n
n factors, so split into n bits A1
(x + a ) A2
(x + a )2 A3
(x + a )3 An
(x + a )n = + + +

12. Numerator same degree or higher than the denominator
Partial Fractions
Numerator same degree or higher than the denominator
Worked example

13. Find partial fractions for
x( x + 3 )
x2 + x - 12
Find partial fractions for
Numerator is degree 2, denominator is degree 2.
Same degree => DIVIDE OUT first 1
QUOTIENT
x2 + x - 12
x2 + 3x + 0
x2 + x - 12
REMAINDER
2x + 12
NOW FIND PARTIAL FRACTIONS FOR THIS BIT
x( x + 3 )
x2 + x - 12
2 x + 12
x2 + x - 12
=> = + 1
THE FINAL ANSWER IS
x( x + 3 )
x2 + x - 12 18
7( x – 3 ) 4
7( x + 4 ) = + - 1

14. Let’s do some more examples
Express
3x2 + 2x + 1
( x + 1 )(x2 + 2x + 2)
in partial fractions.
We must first use the discriminant to verify that the
quadratic factor (x2 + 2x + 2) is irreducible.
For x2 + 2x + 2 :
a = , b = , c =
b2 – 4ac =
22 – 4(1)(2) = -4
b2 – 4ac < 0
=> x2 + 2x + 2 is irreducible
3x2 + 2x + 1
( x + 1 )(x2 + 2x + 2) =

15. as the sum of a polynomial and partial fractions.
Another example:
Express
x3 - 3x
(x2 - x - 2)
as the sum of a polynomial
and partial fractions.