Help needed for Partial Fraction Decomposition

Question

The problem I need help with is this:

$$\frac{x^4} {(x-1)^3}$$

I've already put it into the form: $$ \frac{x^4} {(x-1)^3}=\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}$$

Then, I multiplied the whole equation by the LCD ($(x-1)^3$) then got: $$x^4=A(x-1)^2+B(x-1)+C$$

I let x=1 to find C, which resulted in C being 1.

At this point I tried to replace C with 1, then expand the equation to get: $$x^4=Ax^2-2Ax+A+Bx-B+1$$

From here I don't know how to get the answer, which is $$x+3+\frac{6}{x-1}+\frac{4}{(x-1)^2}+\frac{1}{(x-1)^3}$$

Any help would be much appreciated!

Answer 1

You can't use partial fraction decomposition since the degree of the numerator is greater than the degree of the denominator. Consider using polynomial division first before applying partial fraction decomposition. Given the equation $$\dfrac{x^4}{(x - 1)^3}$$

Expanding the denominator expression, we have $$\dfrac{x^4}{x^3 - 3x^2 + 3x - 1}$$

By polynomial division, we have $$\begin{aligned} \dfrac{x^4 - x(x^3 - 3x^2 + 3x - 1)}{x^3 - 3x^2 + 3x - 1} &= x + \dfrac{x^4 - x^4 + 3x^3 - 3x^2 - x}{x^3 - 3x^2 + 3x - 1}\\ &= x + \dfrac{3x^3 - 3x^2 - x}{x^3 - 3x^2 + 3x - 1}\\ &= x + 3 + \dfrac{3x^3 - 3x^2 - x - 3(x^3 - 3x^2 + 3x - 1)}{x^3 - 3x^2 + 3x - 1}\\ &= x + 3 + \dfrac{6x^2 - 8x + 3}{x^3 - 3x^2 + 3x - 1} \end{aligned}$$

Finally, as you said before, by partial fraction decomposition (left as exercise for you to apply partial fraction decomposition), you get:

$$x + 3 + \dfrac{6}{x - 1} + \dfrac{4}{(x - 1)^2} + \dfrac{1}{(x - 1)^3}$$

Answer 2

You cannot immediately apply partial fraction decomposition because the numerator has larger degree than the denominator. To proceed, you could divide them, or you could first write $\,x^3/(x\!-\!1)^3$ as a partial fraction, then multiply it by $\,x = (x\!-\!1) + 1.\,$ But there is an easier way. Change variables: $ $ let $\ t = x\!-\!1,\ $ hence

$$ \dfrac{x^4}{(x\!-\!1)^3}\, =\, \dfrac{(t\!+\!1)^4}{t^3}\, =\, \dfrac{(t^4\! + 4 t^3\!+6t^2\!+4t+1}{t^3}\, =\, t +4+\dfrac{6}{t}+\dfrac{4}{t^2}+\dfrac{1}{t^3}$$

Finally, substituting $\, t = x\!-\!1\,$ yields the sought partial fraction expansion.