# How do you evaluate \frac { ( x + 1) ^ { 2} } { ( x - 2) ( x - 1) ^ { 2} } = \frac { A } { x - 2} + \frac {B}{x-1}+\frac { c } { ( x - 1) ^ { 2} }?

Aug 1, 2017

The answers are $A = 9 ,$B=-8$\mathmr{and}$C=-4#

#### Explanation:

${\left(x + 1\right)}^{2} / \left(\left(x - 2\right) {\left(x - 1\right)}^{2}\right) = \frac{{x}^{2} + 2 x + 1}{\left(x - 2\right) {\left(x - 1\right)}^{2}}$

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

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

The denominators are the same, we compare the numerators

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

Let $x = 1$, $\implies$, $4 = - C$, $\implies$, $C = - 4$

Let $x = 2$, $\implies$, $9 = A$

Coefficient of ${x}^{2}$

$1 = A + B$, $\implies$, $B = 1 - A = 1 - 9 = - 8$

Therefore,

${\left(x + 1\right)}^{2} / \left(\left(x - 2\right) {\left(x - 1\right)}^{2}\right) = \frac{9}{x - 2} - \frac{8}{x - 1} - \frac{4}{x - 1} ^ 2$