Question

Resolve the following rational expressions into partial fractions

`(x^3 + 2x + 1)/(x^2 + 5x + 6)`

Answer 1

`(x^3 + 2x + 1)/(x^2 + 5x + 6)`

Since the numerator is of degree greater than that of the denominator divide the numerator by the denominator.

∴ `(x^3 + 2x + 1)/(x^2 + 5x + 6) = (x - 5) + (21x + 31)/(x^2 + 5x + 6)`  ......(1)

Consider `(21x + 31)/(x^2 + 5x + 6)`

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

= `(21x + 31)/(x(x + 3) + 2(x + 3))`

= `(21x + 31)/((x + 2)(x + 3))`

`(21x + 31)/(x^2 + 5x + 6) = "A"/(x + 2) + "B"/(x + 3)`  ......(2)

`(21x + 31)/(x^2 + 5x + 6) = ("A"(x + 3) + "B"(x + 2))/((x + 2)(x + 3))`

21x + 31 = A(x + 3) + B(x + 2)   ......(3)

Put x = – 3, in equation (3)

21(– 3) + 31 = A(– 3 + 3) + B(– 3 + 2)

– 63 + 31 = 0 – B

– 32 = – B

⇒ B = 32

Put x = – 2, in equation (3)

21(– 2) + 31 = A(– 2 + 3) + B(– 2 + 2)

– 42 + 31 = A + 0

⇒ A = – 11

Substituting the values of A and B in equation (2)

We have `(21x + 31)/(x^2 + 5x + 6) = (-11)/(x + 2) + 32/(x + 3)`

= `- 11/(x + 2) + 32/(x + 3)`

∴ The required partial fraction is

`(x^3 + 2x + 1)/(x^2 + 5x + 6) = (x - 5) - 11/(x + 2) + 32/(x + 3)`