# PARTIAL FRACTION WITH CUBIC EQUATION IN THE DENOMINATOR

Partial Fraction with Cubic Equation in the Denominator :

Here we are going to see some practice questions on partial fractions with cubic equation in the denominator.

## Partial Fraction with Cubic Equation in the Denominator - Practice Questions with Answer

Question 1 :

Resolve the following rational expression into partial fractions.

Solution :

Let us decompose the denominator into linear factors.

(6x2 - x + 1)/(x3 + x2 + x + 1)

=  A/(x + 1) + (Bx + C)/(x2 + 1)

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

Coefficients of x2, x and constants

6  =  A + B  -----(1)

-1  =  B + C  -----(2)

1  =  A + C  -----(3)

 (1) - (2)A + B - (B + C)  =  6 + 1A - C  =  7  -----(4) (3) + (4)A + C + A - C  =  1 + 72A  =  8A  =  8/2  =  4

By applying the value of A in the 3rd equation

4 + C  =  1

C  =  1 - 4  =  -3

C  =  -3

By applying the value of B in the 2nd equation

B + C  =  -1

B - 3  =  -1

B  =  -1 + 3

B  =  2

Question 2 :

Resolve the following rational expression into partial fractions.

Solution :

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

(x - 5)/(x + 3) (x - 1)  =  A/(x + 3) + B/(x - 1)

x - 5  = A(x - 1) + B(x + 3)

 When x  =  11 - 5  =  B (1 + 3)-4  =  4BB  =  -1 When x  =  -3-3 - 5  =  A (-3 - 1)-8  =  -4AA  =  2

Question 3 :

Resolve the following rational expression into partial fractions.

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

Solution :

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

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

Equating the coefficients of x2, x and constants.

0  =  A + B  -----(1)

1  =  B + C  -----(2)

7  =  A + C -----(3)

 (1) - (2)A + B - (B + C)  =  0 - 1A + B - B - C  =  - 1A - C  =  -1  ----(4) (3) + (4)A + C + A - C  =  7 - 12A  =  6A  =  3

By applying the value of A in the first equation, we get

3 + B  =  0

B  =  -3

By applying the value of B in the second equation, we get+

-3 + C  =  1

C  =  1 + 3

C  =  4

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

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

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