PARTIAL FRACTIONS QUADRATIC NUMERATOR AND DENOMINATOR

Example 1 :

Resolve the following rational expression into partial fractions.

(x2 + x + 1)/(x2 - 5x + 6)

Solution :

Since we have same power for both numerator and denominator, we have use long division.

(x2 + x + 1)/(x2 - 5x + 6)  =  1 + [(6x - 5)/(x2 - 5x + 6)]

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

(6x - 5)/(x - 2) (x - 3)  =  A/(x - 2) + B/(x - 3)

6x - 5  =  A(x - 3) +  B(x - 2)

If x  =  2

12 - 5  =  A(2 - 3)

7  =  -A

A  =  -7

If x  =  3

18 - 5  =  B(3 - 2)

13  =  B

B  =  13

Hence the solution is

Example 2 :

Resolve the following rational expression into partial fractions.

(x3 + 2x + 1)/(x2 + 5x + 6)

Solution :

The denominator is having least power than numerator, then we have to used long division.

21x + 31/(x2 + 5x + 6)  =  21x + 31/(x + 2)(x + 3)

(21x + 31)/(x2 + 5x + 6)  =  A/(x + 2) + B/(x + 3)

21x + 31  =  A(x + 3) + B(x + 2)

If x  =  -3

-63 + 31  =  B(-3 + 2)

-32  =  - B

B  =  32

If x  =  -2

-42 + 31  =  A(-2 + 3)

-11  =  A

A  =  -11

Hence the solution is

Example 3 :

Resolve the following rational expression into partial fractions.

(x + 12) / (x + 1)2 (x - 2)

Solution :

The denominator is having least power than numerator, then we have to used long division.

x + 12  =  A(x + 1)(x - 2) + B(x - 2) + C(x + 1)2

x + 12  =  A(x2 - x - 2) + B(x - 2) + C(x2 + 2x + 1)

x + 12  =  (A + C)x2 + (-A + B + 2C)x + (-2A - 2B + C)

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

A + C  =  0  ---(1)

-A + B + 2C  =  1   ---(2)

-2A - 2B + C  =  12  ---(3)

(2) ⋅ 2 +  (3) ===>

-2A + 4C -2A + C  =   2 + 12

-4A + 5C  =  14  ---(4)

(1) ⋅ 4 +  (4) -->

 4A + 4C  =  0

-4A + 5C  =  14

-----------------

9C  =  14

C  =  14/9

Applying the value of C in the 1st equation, we get

A + (14/9)  =  0

A  =  -14/9

By applying the value of A and C in the second equation.

(14/9) + B + 2(14/9)  =  1  

B  =  1 - (14/9) - (28/9)

B  =  (9 - 14 - 28)/9

B  =  -33/9

B  =  -11/3

Hence the solution is

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