May 25th

May 25th

Simple Partial Fractions

The last quiz was an example of partial fractions , i.e. the technique of decomposing a fraction as a sum of simpler fractions. This section will consider the simpler forms of this technique.

Example

Find the partial fraction decomposition of 4 / ( x - 4) .

Solution

The denominator factorises as x - 4 = ( x - 2)( x +2) . (See the package on quadratics .) The partial fractions will, therefore, be of the form a/ ( x - 2) and b/ ( x + 2) . Thus

so that ( a + b ) x + 2( a - b ) = 4

The last line is

( a + b ) x + 2( a - b ) = 4 ,

and this enables a and b to be found. For the equation to be true for all values of x the coeffcients must match, i.e.

a + b = 0 (coeffcients of x )

2 a - 2 b = 4 (constant terms)

where the first equation holds since there is no x term in 4 / ( x - 4). This set of simultaneous equations may be solved to give a = 1 and b = - 1. Thus

Exercise

Find a and b .

Solution

Taking common denominators:

so that ( a + b ) x + (2 a - 2 b ) = 4 x. Equating coeffcients in this case gives

a + b = 4 (coeffcients of x )

2 a - 2 b = 0 (constant terms)

Solving this set of equations gives a = 2 , b = 2. Hence

Quiz

If which of the following is the solution to the equation?

Solution

Writing all the fractions with a common denominator

so that (3 a + 2 b ) x + (4 a - 3 b ) = x + 7. This gives two equations

3 a + 2 b = 1 (coeffcients of x )

4 a - 3 b = 7 (constant terms)

Solving these gives a = 1 , b = - 1.