2 Jan 1999
efinitions
When a given rational function f(x)/g(x), where
f(x) and g(x) are polynomials in x, is expressed as the sum of two or more simpler fractions according to certain
rules, it is said to be expressed in partial fractions.
A rational function f(x)/g(x) is proper
if deg f(x) < deg g(x). Otherwise
it is called an improper fraction.
If f(x)/g(x) is improper,
then by long division
|
f(x) |
= g(x)Q(x)
+ R(x) |
|
|
\
|
f(x)
¾¾
g(x) |
|
where R(x)/g(x) is
a proper fraction. |
ules For Partial Fractions
Before expressing a rational function into partial fractions, the following steps should be performed:
- f(x), g(x) are factorised to eliminate common factors.
- If f(x)/g(x) is improper, it has to be broken down as Q(x) + R(x)/g(x).
- g(x) must be completely factorised into its linear or irreducible quadratic factors.
If a rational function f(x)/g(x) is a proper fraction, we can express it into partial fractions according to the following rules.
-
Every non-repeated linear factor (ax + b) in g(x) corresponds to a fraction of the form:
- Every linear factor (ax + b) in g(x) that is repeated n times corresponds to a sum of n partial fractions:
A1
¾¾¾
ax + b
|
+ |
A2
¾¾¾¾
(ax + b)2
|
+ ¼ + |
An
¾¾¾¾
(ax + b)n
|
. |
-
Every non-repeated irreducible quadratic factor (ax2 + bx + c) in g(x) corresponds to a fraction of the form:
Ax + B
¾¾¾¾¾
ax2 + bx + c
|
. |
ethods Of Solving Unknowns
- By substitution
- By comparing coefficients
- By cover-up method
xamples
Example 1:
x - 9
|
|
A
|
|
B
|
¾¾¾¾¾¾ |
= |
¾¾¾ |
+ |
¾¾¾ |
. |
(x - 1)(x + 3)
|
|
x - 1
|
|
x + 3
|
By the 'cover-up' method,
|
1 - 9
|
|
-3 - 9
|
|
A = |
¾¾¾ |
= -2 and B
= |
¾¾¾ |
= 3. |
|
1 + 3
|
|
-3 - 1
|
|
|
x - 9
|
|
-2
|
|
3
|
Hence |
¾¾¾¾¾¾ |
= |
¾¾¾ |
+ |
¾¾¾ |
. |
|
(x - 1)(x + 3)
|
|
x - 1
|
|
x + 3
|
Example 2:
x2 + 6x + 9
|
|
A
|
|
B
|
|
C
|
¾¾¾¾¾¾ |
= |
¾¾¾ |
+ |
¾¾¾ |
+ |
¾¾¾ |
. |
(x - 3)2(x
+ 5)
|
|
x - 3
|
|
(x - 3)2
|
|
x + 5
|
Then x2 + 6x + 9 = A(x - 3)(x + 5) + B(x + 5) + C(x - 3)2.
Put x = 3, we get 36 = B(8)
Þ B = 9/2.
Put x = - 5,
we get 4 = C(64) Þ
C = 1/16.
Equating coefficient of x2,
we have 1 = A + C Þ
A = 15/16.
|
x2 + 6x + 9
|
|
15
|
|
9
|
|
1
|
Hence |
¾¾¾¾¾¾ |
= |
¾¾¾¾ |
+ |
¾¾¾¾ |
+ |
¾¾¾¾ |
. |
|
(x - 3)2(x
+ 5)
|
|
16(x - 3)
|
|
2(x - 3)2
|
|
16(x + 5)
|
Example 3:
x3 + 3x
|
¾¾¾¾¾¾¾ |
. |
(x - 1)(2x2
+ 3)
|
Note that this is an improper fraction.
\ We have to carry
out the long division before expressing into partial fractions.
x3 + 3x
|
¾¾¾¾¾¾¾ |
(x - 1)(2x2
+ 3)
|
|
= |
1
|
|
x2 + (3/2)x + (3/2)
|
¾ |
+ |
¾¾¾¾¾¾¾¾ |
2
|
|
(x - 1)(2x2
+ 3)
|
|
|
|
= |
1
|
|
A
|
|
Bx + C
|
¾ |
+ |
¾¾¾ |
+ |
¾¾¾¾ |
. |
2
|
|
x - 1
|
|
2x2 + 3
|
|
Then x2 + (3/2)x + (3/2) = A(2x2 + 3) + (Bx + C)(x - 1).
Put x = 1, 4 = 5A Þ A = 4/5.
Compare coefficient of x2, 1
= 2A + B Þ B = -3/5.
Compare constant term, 3/2 = 3A - C Þ C = 9/10.
|
x3 + 3x
|
|
1
|
|
4
|
|
6x - 9
|
Hence |
¾¾¾¾¾¾¾ |
= |
¾ |
+ |
¾¾¾¾ |
- |
¾¾¾¾¾ |
. |
|
(x - 1)(2x2
+ 3)
|
|
2
|
|
5(x - 1)
|
|
10(2x2 + 3)
|
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