# PARTIAL FRACTION WITH CUBIC DENOMINATOR

Partial Fraction with Cubic Denominator :

Here we are going to see some examples on decomposition of partial fractions with cubic denominator.

## Partial Fraction with Cubic Denominator - Examples

Example 1 :

Resolve the following rational expressions into partial fractions.

(6x2 - x + 1)/(x3 + x2 + x + 1)

Solution :

Let us use synthetic division to factorize the cubic denominator.

6x2 - x + 1  =  A(x2 + 1) + (Bx + C)(x + 1)

6x2 - x + 1  =  Ax2 + A + Bx2 + Bx + Cx + C

Equating the coefficient of x2, x and constant terms

6  =  A + B  ----(1)

-1  =  B + C  ----(2)

1  =  A + C   ----(3)

(1) - (2)

A + B - (B + C)  =  6 + 1

A - C  =  7 ----(4)

(3) + (4)  ==>  A + C + A - C  =  1 + 7

2A  =  8

A  =  4

 Apply A = 4 in (1)4 + B  =  6B  =  6 - 4B  =  2 Apply B = 2 in (2)2 + C  =  -1C  =  - 1 - 2C  =  -3

Hence the solution is

Example 2 :

Resolve the following rational expressions into partial fractions.

(2x2 + 5x - 11)/(x2 + 2x - 3)

Solution :

(2x2 + 5x - 11)/(x2 + 2x - 3)

=  2 + [(x - 5)/(x + 3)(x - 1)]

x - 5  =  A/(x + 3) + B(x - 1)

x - 5  =  A(x - 1) + B(x + 3)

 If x = 1-4  =  B(4)4B  =  -4B  =  -1 If x = -3-8  =  A(-4)A  =  8/4A  =  2

Hence the solution is

Example 3 :

Resolve the following rational expressions into partial fractions.

(7 + x)/(1 + x)(1 + x2)

Solution :

(7 + x)/(x + 1)(x2 + 1)  =  A/(x + 1) + (Bx+C)/(x2 + 1)

7 + x  =  A(x2 + 1) + (Bx + C)(x + 1)

0  =  A + B  ---(1)

1  =  B + C  ---(2)

7  =  A + C ---(3)

(1) - (2)===>

A - C  =  -1  ---(4)

(3) + (4)  ==>

2A  =  6  ===> A  =  3

B  =  -3

C  =  A + 1  ==> 3 + 1 ==> 4

C  =  4

Hence the solution is

3/(x + 1) + (4 - 3x)/(x2 + 1)

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

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