Question

Resolve into partial fractions:
$\frac{9}{(x-1){(x+2)}^2}$

Answer 1

$\frac{9}{(x-1){(x+2)}^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{{(x+2)}^2}$

$=>9=A{(x+2)}^2+B(x-1)(x+2)+C(x-1)$ -(1)

For $x=1,=>9=A{\left(3\right)}^2+B\left(0\right)+C\left(0\right)$

$=>9=9A$

$=>A=1$ - (2)

For $x=-2,=>9=A\left(0\right)+B\left(0\right)+C(-2-1)$

$=>9=-3C$

$=>C=-3$ -(3)

Now , we have found out value of A and C , to find value of B , put x = any real number .

Let x =0 , then equation (1) reduces to $9=4A-2B-C$

$=>9=4-2B$

$=>2=-2B$

$=>B=-1$ -(4)

Therefore from (2),(3),(4) we get $\frac{9}{(x-1){(x+2)}^2}=\frac{1}{x-1}-\frac{1}{x+2}-\frac{3}{{(x+2)}^2}$