 Partial Fraction Decomposition How to solve partial fraction decomposition problems: examples and their solutions.

Example 1 Write the given fraction
as the sum of the fractions
whose denominators are
the reduced factors of the given denominator:

(given) = A/x + B/(x - 1)

Your goal is to find A and B.

To add the fractions,
make each fraction's denominator
to the given fraction's denominator.

Multiply the missing factors
to both of the numerator and the denominator.

Adding and subtracting rational expressions

The numerators of both side fractions are equal.

So 3x - 2 = A(x - 1) + Bx.

To compare the terms of both sides,
arrange the right side in descending order.

Ascending order, descending order

Then switch both sides.

(A + B)x - A = 3x - 2

Compare the constant terms on both sides:
-A = -2.

Then A = 2.

Compare the x terms' coefficients on both sides:
A + B = 3.

A = 2
So 2 + B = 3.

Then B = 1.

Put A = 2 and B = 1
into A/x + B/(x - 1).

Then
(given) = 2/x + 1/(x - 1).

Example 2 Write the given fraction
as the sum of the fractions
whose denominators are
the reduced factors of the given denominator:

(given) = A/x + B/x2 + C/(x - 1)

Your goal is to find A, B, and C.

To add the fractions,
make each fraction's denominator
to the given fraction's denominator.

Multiply the missing factors
to both of the numerator and the denominator.

Adding and subtracting rational expressions

The numerators of both side fractions are equal.

So 5x2 - 1 = Ax(x - 1) + B(x - 1) + Cx2.

To compare the terms of both sides,
arrange the right side in descending order.

Ascending order, descending order

Then switch both sides.

(A + C)x2 + (A + B)x + B = 5x2 - 1

Compare the constant terms on both sides:
B = -1.

Compare the x terms' coefficients on both sides:
A + B = 0.

B = -1
So A + -1 = 0.

Then A = 1.

Compare the x2 terms' coefficients on both sides:
A + C = 5.

A = 1
So 1 + C = 5.

Then C = 4.

Put A = 1, B = -1, and C = 4
into A/x + B/x2 + C/(x - 1).

Then
(given) = 1/x - 1/x2 + 4/(x - 1).