2 Proper fractions with linear factors

Firstly we describe how to calculate partial fractions for proper fractions where the denominator may be written as a product of linear factors. The steps are as follows:

∙ Factorise the denominator.

∙ Each factor will produce a partial fraction. A factor such as 3 x + 2 will produce a partial

fraction of the form A 3 x + 2 where A is an unknown constant. In general a linear factor

a x + b will produce a partial fraction A a x + b . The unknown constants for each partial

fraction may be different and so we will call them A , B , C and so on.

∙ Evaluate the unknown constants by equating coefficients or using specific values of x .

The sum of the partial fractions is identical to the original algebraic fraction for all values of x .

Key Point 14

A linear factor a x + b in the denominator gives rise to a single partial fraction of the form  A a x + b

The steps involved in expressing a proper fraction as partial fractions are illustrated in the following Example.

Example 41

Express 7 x + 10 2 x 2 + 5 x + 3 in terms of partial fractions.

Solution

Note that this fraction is proper. The denominator is factorised to give ( 2 x + 3 ) ( x + 1 ) . Each of the linear factors produces a partial fraction. The factor 2 x + 3 produces a partial fraction of the form A 2 x + 3 and the factor x + 1 produces a partial fraction B x + 1 , where A and B are constants which we need to find. We write

7 x + 10 ( 2 x + 3 ) ( x + 1 ) = A 2 x + 3 + B x + 1

 By multiplying both sides by ( 2 x + 3 ) ( x + 1 ) we obtain

7 x + 10 = A ( x + 1 ) + B ( 2 x + 3 ) …(*)

We may now let x take any value we choose . By an appropriate choice we can simplify the right-hand side. Let x = − 1 because this choice eliminates A . We find

7 ( − 1 ) + 10 = A ( 0 ) + B ( − 2 + 3 ) 3 = B

so that the constant B must equal 3. The constant A can be found either by substituting some other value for x or alternatively by ‘equating coefficients’.

Observe that, by rearranging the right-hand side, Equation (*) can be written as

7 x + 10 = ( A + 2 B ) x + ( A + 3 B )

Comparing the coefficients of x on both sides we see that 7 = A + 2 B . We already know B = 3 and so

7 = A + 2 ( 3 ) = A + 6

from which A = 1 . We can therefore write

7 x + 10 2 x 2 + 5 x + 3 = 1 2 x + 3 + 3 x + 1

We have succeeded in expressing the given fraction as the sum of its partial fractions. The result can always be checked by adding the fractions on the right.

Task!

Express 9 − 4 x 3 x 2 − x − 2 in partial fractions.

First factorise the denominator:

( 3 x + 2 ) ( x − 1 )

Because there are two linear factors we write

9 − 4 x 3 x 2 − x − 2 = A 3 x + 2 + B x − 1

 Multiply both sides by ( 3 x + 2 ) ( x − 1 ) to obtain the equation from which to find A and B :

9 − 4 x = A ( x − 1 ) + B ( 3 x + 2 )

Substitute an appropriate value for x to obtain B :

Substitute x = 1 and get B = 1

Equating coefficients of x to obtain the value of A :

− 4 = A + 3 B , A = − 7 since B = 1

Finally, write down the partial fractions:

− 7 3 x + 2 + 1 x − 1

Exercises
  1. Find the partial fractions of
    1. 5 x − 1 ( x + 1 ) ( x − 2 ) ,  
    2. 7 x + 25 ( x + 4 ) ( x + 3 ) ,  
    3. 11 x + 1 ( x − 1 ) ( 2 x + 1 ) .

      Check by adding the partial fractions together again.

  2. Express each of the following as the sum of partial fractions:
    1. 3 ( x + 1 ) ( x + 2 ) ,
    2. 5 x 2 + 7 x + 12 ,
    3. − 3 ( 2 x + 1 ) ( x − 3 ) ,

1(a)   2 x + 1 + 3 x − 2 , 1(b)   3 x + 4 + 4 x + 3 1(c)   4 x − 1 + 3 2 x + 1 ,

2(a)   3 x + 1 − 3 x + 2 , 2(b)   5 x + 3 − 5 x + 4 , 2(c)   6 7 ( 2 x + 1 ) − 3 7 ( x − 3 ) .