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)
