# How do you write the partial fraction decomposition of the rational expression (3x^2 -7x+1) / (x-1)^3?

Nov 21, 2016

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

#### Explanation:

The decomposition in partial fractions is

$\frac{3 {x}^{2} - 7 x + 1}{x - 1} ^ 3 = \frac{A}{x - 1} ^ 3 + \frac{B}{x - 1} ^ 2 + \frac{C}{x - 1}$

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

Therefore,

$\left(3 {x}^{2} - 7 x + 1\right) = A + B \left(x - 1\right) + C {\left(x - 1\right)}^{2}$

Let $x = 1$, $\implies$, $- 3 = A$

$A + B + C = 1$

Coefficients of ${x}^{2}$
$3 = C$

Coefficients of $x$

$- 7 = - B - 2 C$, $\implies$, $B = - 1$

so, $\frac{3 {x}^{2} - 7 x + 1}{x - 1} ^ 3 = \frac{- 3}{x - 1} ^ 3 - \frac{1}{x - 1} ^ 2 + \frac{3}{x - 1}$