# How do you write x^4/(x-1)^3 as a partial fraction decomposition?

Oct 27, 2016

The result is
${x}^{4} / {\left(x - 1\right)}^{3} = x + 3 + \frac{6}{x - 1} + \frac{4}{x - 1} ^ 2 + \frac{1}{x - 1} ^ 3$

#### Explanation:

Since the degree of the numerator is greater than the degree of the denominator, we perform a long division

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

${x}^{4}$$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}$∣${x}^{3} - 3 {x}^{2} + 3 x - 1$

${x}^{4} - 3 {x}^{3} + 3 {x}^{2} - x$ $\textcolor{w h i t e}{a a a a a}$∣$x + 3$

$0 + 3 {x}^{3} - 3 {x}^{2} + x$

$\textcolor{w h i t e}{a a a}$$3 {x}^{3} - 9 {x}^{2} + 9 x - 3$

$\textcolor{w h i t e}{a a a a a}$$0 + 6 {x}^{2} - 8 x + 3$

So we get

${x}^{4} / {\left(x - 1\right)}^{3} = x + 3 + \frac{6 {x}^{2} - 8 x + 3}{x - 1} ^ 3$

now we form the partial fraction decomposition

$\frac{6 {x}^{2} - 8 x + 3}{x - 1} ^ 3 = \frac{A}{x - 1} + \frac{B}{x - 1} ^ 2 + \frac{C}{x - 1} ^ 3$

$\frac{6 {x}^{2} - 8 x + 3}{x - 1} ^ 3 = \frac{A {\left(x - 1\right)}^{2} + B \left(x - 1\right) + C}{x - 1} ^ 3$

So $6 {x}^{2} - 8 x + 3 = A {\left(x - 1\right)}^{2} + B \left(x - 1\right) + C$

Let $x = 1$ then $1 = C$

Compare the coefficients of ${x}^{2}$

$6 = A$

Let $x = 0$, then $3 = A - B + C$

$B = 6 + 1 - 3 = 4$

So the final result is

${x}^{4} / {\left(x - 1\right)}^{3} = x + 3 + \frac{6}{x - 1} + \frac{4}{x - 1} ^ 2 + \frac{1}{x - 1} ^ 3$