Method 14

Resolving into Partial Fractions

This method is based on the sutra " Paravartya Yojayet " which means
' transpose and apply ' .
Explaining the method with the help of an example .
Example 1 : - Resolve the following into partial fractions .
3x + 13
(x + 1) (x + 2)
Answer : -
Write the above term as
3.x + 13 = A + B
(x + 1) (x + 2) (x + 1) (x + 2)
Step 1 : -
Take one term in the denominator . Like (x = 1) & obtain the value of x by placing it to 0
i.e. x + 1 = 0 => x = -1
Place this in the remaining term
i.e.
3.x + 13
(x + 2)
= 3.(-1) + 13
(-1 + 2)
= 10
which is the value of A

Step 2 : -

Again place the other term to 0 and obtain the value of x .
x = -2
Place this in the remaining terms
i.e.
3.(-2) + 13
(-2 + 1)
= - 7
which is the value of B .

The answer to the above quuestion is

3.x + 13 = 10 - 7
(x + 1) (x + 2) (x + 1) (x + 2)

Example 2 : - Resolve the following into partial fractions
9
(x² + x - 2)

Answer : -

9 = 9
(x² + x - 2) (x - 1) (x + 2)
9 = A + B
(x -1) (x + 2) (x -1) (x + 2)

Step1 : -

x = 1
Substitute in the remaining term
9 = 9 = 3
(x + 2) 1 + 2
which is the value of A.

Step 2 : -
x = -2
Substitute in the remaining term
9 = 9 = - 3
(x - 1) -2 -1
which is the value of B .

The answer to the above question is
9 = 3 - 3
(x² + x - 2) (x - 1) (x + 2)

Algebraic Proof : -

Taking
x + 3 = x + 3
(x² + x - 2) (x - 1) (x + 2)
x + 3 = A + B
(x - 1) (x + 2) (x - 1) (x + 2)
=> x + 3 = A(x + 2) + B(x - 1)
(x - 1) (x + 2) ( x -1) (x + 2)
=> x + 3 = A(x + 2) + B(x - 1)
Now putting x - 1 = 0 (which is the term below A)we get
A = (x + 3)
(x + 2)

Similarly putting x + 2 = 0 (which is the term below B) we get
B = (x + 3)
(x - 1)
We use the same things in the various steps .

Exercises : -

Resolve the following into partial fractions

1.)

x² + 5x + 3
(x + 2)(x - 7)(x - 4)

2.)
8x² + 9x + 11

          (x - 12)(x + 13)(x + 14) 

3.)       

     2.x -5      
(x - 2) (x - 3)

4.) 

    2x + 7     
(x + 3) (x + 4)

5.)
3x + 4
3x2 + 3x + 2

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