PARTIAL FRACTIONS QUADRATIC NUMERATOR AND DENOMINATOR

Partial fractions quadratic numerator and denominator :

Here we are going to see some examples on decomposition of rational expressions with quadratic numerator and denominator into partial fractions.

Partial fractions Quadratic Numerator and Denominator - Examples

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

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

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