PARTIAL FRACTIONS EXAMPLES

Partial Fractions Examples :

Here we are going to see some example problems on partial fractions.

Partial Fractions Examples

Example 1 :

Resolve into partial fractions.

Solution :

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

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

Since we have same denominators on both side, we can equate the numerators.

1  =  A(x + 1) + B(x - 1)  ------(1)

The constants A and B can also be found by successively giving suitable values for x.

To find A, put x = 1 in (1)

To find B, put x = -1 in (1)

Example 2 :

Resolve into partial fractions.

Solution :

7x - 1/(x²-5x + 6)  =  7x - 1/(x - 3) (x - 2)

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

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

Since we have same denominators on both side, we can equate the numerators.

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

The constants A and B can also be found by successively giving suitable values for x.

To find A, put x = 3 in (1)

To find B, put x = 2 in (1)

Example 3 :

Resolve into partial fractions.

Solution :

(x²+ x + 1)/(x-1)(x-2)(x-3) = A/(x-1) + B/(x-2) + C/(x-3)

Taking L.C.M

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

Since we have same denominators on both side, we can equate the numerators.

x²+ x + 1  =  A(x-2)(x-3) + B(x-1)(x-3) + C(x-2)(x-1) ----(1)

The constants A, B and C can also be found by successively giving suitable values for x.

To find A, put x = 1 in (1)

To find B, put x = 2 in (1)

To find C, put x = 3 in (1)

Example 4 :

Resolve into partial fractions.

Solution :

1/(x-1)(x+2)² = A/(x-1) + B/(x+2) + C/(x+2)²

Taking L.C.M

1/(x-1)(x+2)² = A(x+2)² + B(x-1)(x+2) + C(x-1)/(x-1)(x+2)²

Since we have same denominators on both side, we can equate the numerators.

1 = A(x+2)² + B(x-1)(x+2) + C(x-1)  ----(1)

By equating the coefficients of x², x and constant terms respectively, we get 

A + B = 0  -------(2)

4A + B + C  = 0   -------(3)

4A - 2B - C  =  1  -------(4)

By adding the equations (3) and (4) we get,

00000000000004A + B + C  = 000000000000000000000

00000000000004A - 2B - C  =  10000000000000000000

000000000000-------------------0000000000000000000

0000000000008A - B  =  1------(5)00000000000000000

(5) + (2)

A + B + 8A - B = 0 + 1

9A  =  1 ==> A = 1/9

Example 5 :

Resolve into partial fractions.

Solution :

(x - 2)/(x+2)(x-1)² = A/(x + 2) + B/(x - 1) + C/(x - 1)²

Taking L.C.M

(x-2)/(x+2)(x-1)² = A(x-1)²+B(x-1)(x+2)+C(x+2)/(x+2)(x-1)²

Since we have same denominators on both side, we can equate the numerators.

x - 2 = A(x - 1)² + B(x - 1)(x + 2) + C(x + 2)  --(1)

By equating the coefficients of x², x and constant terms respectively, we get 

A + B = 0  -------(2)

-2A + B + C  = 1   -------(3)

A - 2B + 2C  =  -2  -------(4)

2 x (3) - (4)

2(-2A + B + C) - (A - 2B + 2C)  =  2 - (-2)

-4A - A + 2B + 2B + 2C - 2C  = 2 + 2

-5A + 4B  =  4  -------(5)

5 x (2) + (5)

5A + 5B + (-5A + 4B)  = 0 + 4

5A + 5B -5A + 4B  = 4

9B  =  4  ==> B = 4/9

After having gone through the stuff given above, we hope that the students would have understood, how to decompose a rational expression into partial fractions. 

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