Question

Decompose

(i)x^4-x^3-2x^2+4x+1/x(x-1)^2

(5)

into partial fractions (show all the steps)

Answer 1

First, let's detach the whole part of the fraction:

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

Second, decompose the fraction $\frac{-2x^2+4x+1}{x(x-1)^2}$ to partial fractions:

$\frac{-2x^2+4x+1}{x(x-1)^2}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}$

Let's multiply both parts of this equation by $x(x-1)^2$ and equate the coefficients at equal powers of x:

$-2x^2+4x+1=A(x-1)^2+Bx(x-1)+\\ +Cx=Ax^2-2Ax+A+Bx^2-Bx+\\ +Cx=(A+B)x^2+\\ +(-2A-B+C)x+A\\ \begin{cases} A+B=-2\\ -2A-B+C=4\\ A=1 \end{cases}\Rarr\\ \Rarr\begin{cases} A=1\\ B=-2-A=-2-1=-3\\ -B+C=4+2A=4+2\cdot1=6 \end{cases}\Rarr\\ \Rarr\begin{cases} A=1\\ B=-3\\ C=6+B=6+(-3)=3 \end{cases}$

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