Question

Q:Resolve this into partial fraction.
9/(X-1)(X+2)^2

Answer 1

Answer:

$\frac{1}{x-1}$ - $\frac{1}{x+2}$ - $\frac{3}{(x+2)^2}$

Step-by-step explanation:

Expressing as a partial fraction

$\frac{9}{(x-1)(x+2)^2}$ = $\frac{A}{(x-1)}$ + $\frac{B}{(x+2)}$ + $\frac{C}{(x+2)^2}$ ( A, B, C are numerical values to be found )

Multiply both sides by (x - 1)(x + 2)²

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

Choose value for x that are zeros of the factors and substitute into right side

x = - 2 : 9 = - 3C ⇒ C = - 3

x = 1 : 9 = 9A ⇒ A = 1

x = 0 : 9 = 4A - 2B - C , that is

9 = 4 - 2B + 3 = - 2B + 7 ( subtract 7 from both sides )

2 = - 2B ⇒ B = - 1, then

A = 1, B = - 1, C = - 3

Thus

$\frac{9}{(x-1)(x+2)^2}$ = $\frac{1}{x-1}$ - $\frac{1}{x+2}$ - $\frac{3}{(x+2)^2}$