Find the partial fraction: $\frac{x^2}{(x-1)^2(x-2)^2(x-3)}$

Question

I'm trying to find the partial fraction for the following $\frac{x^2}{(x-1)^2(x-2)^2(x-3)}$ by the process of long-division, here's what I have tried:

Leaving $(x-3)$ as it is in the denominator and aiming for $(x-1)$ and $(x-2)$.

First step: $\frac{1^2}{(x-1)^2(1-2)^2(1-3)} = \frac{1}{-2(x-1)^2}, \frac{2^2}{(2-1)^2(x-2)^2(2-3)} = \frac{4}{-(x-2)^2}$

We should then substract this from the first equation to get:

$\frac{x^2}{(x-1)^2(x-2)^2(x-3)} - \frac{1}{-2(x-1)^2} - \frac{4}{-(x-2)^2} = \frac{9x^3-36-45x^2+72x}{2\left(x-1\right)^2\left(x-2\right)^2\left(x-3\right)}$

Taking synthetic division on the numerator I find that 2 and 3 is not a root as there exists remainders. Have I approached this step in the wrong way?

Answer 1

You've made some error in your simplification.

$$ \frac{x^2}{(x-1)^2(x-2)^2(x-3)} - \frac{1}{-2(x-1)^2} - \frac{4}{-(x-2)^2}$$ $$ \begin{align*} &= \frac{x^2+\frac12(x-2)^2(x-3)+4(x-1)^2(x-3)}{(x-1)^2(x-2)^2(x-3)} \\ &= \frac{9x^3-45x^2+72x-36}{2(x-1)^2(x-2)^2(x-3)} \\[1ex] &= \frac{9(x^3-5x^2+8x-4)}{2(x-1)^2(x-2)^2(x-3)} \\[1ex] &= \frac{9(x-1)(x-2)^2}{2(x-1)^2(x-2)^2(x-3)} \\[1ex] &= \frac{9}{2(x-1)(x-3)} \end{align*}$$

Answer 2

Hint: $f(x) = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x-2}+\dfrac{D}{(x-2)^2}+\dfrac{E}{x-3}$. And you have to find these constants $A,B,C,D,E$.