Partial Fractions
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)
(putting the fractions over a common denominator)
5(x + 2) º A(x + 6) + B(x + 1) (we have cancelled the denominators)
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 any values of x into both sides of the expression to help us find A and B. When trying to work out these constants, try to choose values of x which will make the arithmetic easier. In this example, if we substitute x = -6 into the identity, the A(x + 6) term will disappear, making it much easier to solve.
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) |
Cover Up Method
The "cover-up method" is a quick way of working out partial fractions, but it is important to realise that this only works when there are linear factors in the denominator, as there are here.
To put 5(x + 2) into partial fractions using the cover up method:
(x + 1)(x + 6)
cover up the x + 6 with your hand and substitute -6 into what's left, giving 5(-6 + 2)/(-6+1) = -20/-5 = 4. This tells you that one of the partial fractions is 4/(x + 6). Now cover up (x + 1) and substitute -1 into what's left to discover that the other partial fraction is 1/(x + 1) .
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)2, 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 |
Note that we have put a (x - 1) and a (x - 1)2 fraction in.
As before, all we do now is find the values of A, B and C, by putting them over a common denominator and then substituting in values for x.
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 = -3/4
let x = 0
-2 = A - B + C
-2 = -3/4 - B -½
B = 3/4
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.
