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)
 = Q(x) + R(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:

1. f(x), g(x) are factorised to eliminate common factors.
2. If f(x)/g(x) is improper, it has to be broken down as Q(x) + R(x)/g(x).
3. 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.

1. Every non-repeated linear factor (ax + b) in g(x) corresponds to a fraction of the form:
2.  A ¾¾¾ ax + b .

3. Every linear factor (ax + b) in g(x) that is repeated n times corresponds to a sum of n partial fractions:
4.  A1 ¾¾¾ ax + b + A2 ¾¾¾¾ (ax + b)2 + ¼ + An ¾¾¾¾ (ax + b)n .

5. Every non-repeated irreducible quadratic factor (ax2 + bx + c) in g(x) corresponds to a fraction of the form:
6.  Ax + B ¾¾¾¾¾ ax2 + bx + c .

### ethods Of Solving Unknowns

1. By substitution
2. By comparing coefficients
3. 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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