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)

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)

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)

Hence the solution is

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