Question

Resolve the following rational expressions into partial fractions

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

Answer 1

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

= `(6x^2 - x + 1)/((x^2 + 1)(x + 1))`

`(6x^2 - x + 1)/(x^3 + x^2 + x + 1) = ("A"x + "B")/(x^2 + 1) + "C"/(x + 1)`

`(6x^2 - x + 1)/(x^3 + x^2 + x + 1) = (("A"x + "B")(x + 1) + "C"(x^2 + 1))/((x^2 + 1)(x + 1))`

6x² – x + 1 = Ax (x + 1) + B (x + 1) + C(x² + 1)  ......(1)

Put x = – 1 in equation (1)

6 x (– 1)² – (– 1) + 1 = A(– 1 )(– 1 + 1) + B(– 1 + 1) + C( (– 1)² + 1)

6 + 1 + 1 = A × 0 + B × 0 + C (2)

8 = 2C

⇒ C = 4

Put x = 0 in equation (1)

6 × 0² – 0 + 1 = A(0)(0 + 1) + B(0 + 1) + C(0² + 1)

1 = 0 + B + C

1 = B + 4

B = 1 – 4 = – 3

Equating the coefficient of x² in equation (1) we have

6 = A + C

6 = A + 4

⇒ A = 6 – 4

⇒ A = 2

∴ The required partial fraction is

`(6x^2 - x + 1)/(x^3 + x^2 + x + 1) = (2x - 3)/(x^2 + 1) + 4/(x + 1)`