Partial Fractions
Bookmark this page

It is possible to split many fractions into the sum or difference of two or more fractions. This has many uses (such as in integration).
At GCSE level, we saw how:

  1     +     4        =       5(x + 2)    
(x + 1)      (x + 6)          (x + 1)(x + 6)

The method of partial fractions allows us to split the right hand side of the above equation into the left hand side.

Linear Factors in Denominator
This method is used when the factors in the denominator of the fraction are linear (in other words do not have any square or cube terms etc).

Example:

Split    5(x + 2)      into partial fractions.
       (x + 1)(x + 6)

We can write this as:

  5(x + 2)      º    A       +     B  
(x + 1)(x + 6)      (x + 1)       (x + 6)

So now, all we have to do is find A and B.

\   5(x + 2)      º    A(x + 6) + B(x + 1)
   (x + 1)(x + 6)             (x + 1)(x + 6)

\  5(x + 2) º A(x + 6) + B(x + 1)

The above expression is an identity (hence º rather than =). An identity is true for every value of x. This means that we can substitute values of x into both sides of the expression to help us find A and B.

when x = -6,
   5(-4)  =  B(-5)
\ B = 4

when x = -1,
  5(1)  =  5A
\ A = 1

since      5(x + 2)      º    A       +     B  
         (x + 1)(x + 6)      (x + 1)       (x + 6)

the answer is     1         +      4     (as we knew)
                     (x + 1)        (x + 6)

Repeated Factor in the Denominator
Remember, the above method is only for linear factors in the denominator. When there is a repeated factor in the denominator, such as (x - 1)² or (x + 4)³, the following method is used.

Example:
Split         x  -  2         into partial fractions
         (x + 1)(x - 1)²

This time we write:

      x - 2         º          A       +      B       +      C    
(x + 1)(x - 1)²         (x + 1)        (x - 1)         (x - 1)²

As before, all we do now is find the values of A, B and C.
x - 2 º A(x - 1)² + B(x - 1)(x + 1) + C(x + 1)

let x = 1
-1 = 2C
C = -½

let x = -1
-3 = 4A
A = -¾

let x = 0
-2 = A - B + C
-2 = -¾ - B -½
B = ¾

Therefore the answer is:
  - 3      +       3       -      1      
4(x + 1)      4(x - 1)      2(x - 1)²

Quadratic Factor in the Denominator
This method is for when there is a square term in one of the factors of the denominator.

Example:
     2x - 1         º        A      +    Bx + C
(x + 1)(x² + 1)       (x + 1)        (x² + 1)

Find A, B and C in the same way as above.

Note that it is Bx + C on the numerator of the fraction with the squared term in the denominator.

© Matthew Pinkey

Other Notes in this Category

1. Algebraic Long Division
2. Functions
3. Indicies
4. Logarithms
5. Partial Fractions
6. Reduction to Linear Form
7. Sequences
8. Series
9. Set Theory
10. Simultaneous Equations
11. Surds
12. The Binomial Series

Didn't find this useful?

• Visit Coursework.Info for over 14,000 GCSE, A-Level and University Essays