## PARTIAL FRACTIONS

In this page partial fractions we are going to see how to solve some example problems using the concept partial-fraction.First let us see the definition of partial-fraction.

Definition:

Any proper fraction can be expressed as the sum of other two simple fractions corresponding to the factors of the denominator of the given fractions. This process is known as partial fractions.

Now let us see how a single fraction is being expressed as sum of two fractions.

How to split the given fraction as partial-fraction:

First we have to factorize the denominator into prime factors.There are three types in partial-fraction.

Type 1 :

Linear factors, no factor is repeated

When the factors of the denominator of the given fraction are all linear factors none of which is repeated. We write the partial-fraction as follows.

Example :

(x +3)/(x + 1) (x - 2)

here the denominator is in the form linear factors  and no factor is repeated. So we can write the partial-fraction as

(x +3)/(x + 1) (x - 2) = [A/(x + 1)] + [B/(x - 2)]

where A and B are constants.

Type 2 :

Linear factors, Some of the factors are repeated

If a linear factor (a x + b) occurs n times as a factor of the denominator of the given fraction, then we can write the partial-fraction as

Example :

(x +3)/(x - 2)³ = [A/(x - 2)] + [B/(x - 2)²] + [C/(x - 2)³]

where A, B and C are constants.

Type 3 :

If a quadratic equation a x² + b x + c which is not favorable into linear factors occurs only once as factor of the denominator of the given fraction, then we can write the partial fraction as

Example :

(x +3)/(x - 1) (x² + 2 x - 2) = [A/(x - 1)] + [(B x + c)/(x² + 2 x - 2)]

If we have only one quadratic equation alone in the denominator we can write it as simply (Ax + B)/(the given quadratic equation)

Now we are going to see an example to understand this concept

Example:

Resolve into partial fractions  (3 x + 7)/(x² - 3 x +2)

Solution:

The denominator can be factorized as into linear factors

x² - 3 x + 2 = x² - x - 2 x + 2

= x (x - 1) - 2 (x - 1)

= (x - 2) (x - 1)

(3 x + 7)/(x² - 3 x +2) = (3 x + 7)/(x - 2) (x - 1)

(3 x + 7)/(x - 2) (x - 1) =  [A/(x - 2)] + [B/(x - 1)]

3 x + 7 = A (x - 1) + B (x - 2)

Now we have to find the values of the constants A and B. For that we have to give some values for x. That is to find the value of the constant B we have to make the constant A as zero for that we are going to plug 1 for x

plug x = 1

3 (1) + 7 = A (1 - 1) + B (1 - 2)

3 + 7 = A (0) + B (- 1)

10 = B (- 1)

B = -10

Now we are going to plug 2 for x.

plug x = 2

3 (2) + 7 = A (2 - 1) + B (2 - 2)

6 + 7 = A (1) + B (0)

13 = A (1) + 0

13 = A

(3 x + 7)/(x² - 3 x +2) = [13/(x - 2)] + [-10/(x - 1)]

= [13/(x - 2)] - [10/(x - 1)]

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