# How do you express (x^2-2x+1)/(x-2)^3 in partial fractions?

Aug 7, 2018

The answer is $= \frac{1}{x - 2} ^ 3 + \frac{2}{x - 2} ^ 2 + \frac{1}{x - 2}$

#### Explanation:

Perform the decomposition into partial fractions as follows

$\frac{{x}^{2} - 2 x + 1}{x - 2} ^ 3 = \frac{A}{x - 2} ^ 3 + \frac{B}{x - 2} ^ 2 + \frac{C}{x - 2}$

$= \frac{A + B \left(x - 2\right) + C {\left(x - 2\right)}^{2}}{{\left(x - 2\right)}^{3}}$

The denominators are the same, compare the numerators

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

Let $x = 2$, $\implies$, $4 - 4 + 1 = A$, $\implies$, $A = 1$

Coefficients of ${x}^{2}$

$1 = C$

Coefficients of $x$

$- 2 = B - 4 C$

$B = 4 C - 2 = 2$

Finally,

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