Answer
$\dfrac{x^5}{x^2-4x+4}=x^3+4x^2+12x+32+\dfrac{80}{x-2}+\dfrac{288}{(x-2)^2}$
Work Step by Step
We are given the fraction:
$\dfrac{x^5}{x^2-4x+4}$
Because the degree of the numerator is greater than the degree of the denominator, we divide the numerator by the denominator:
$\dfrac{x^5}{x^2-4x+4}=\dfrac{x^3(x^2-4x+4)+4x^4-4x^3}{x^2-4x+4}=\dfrac{x^3(x^2-4x+4)+4x^2(x^2-4x+4)+12x^3-16x^2}{x^2-4x+4}$
$=\dfrac{x^3(x^2-4x+4)+4x^2(x^2-4x+4)+12x(x^2-4x+4)+32x^2-48x}{x^2-4x+4}$
$=\dfrac{x^3(x^2-4x+4)+4x^2(x^2-4x+4)+12x(x^2-4x+4)+32(x^2-4x+4)+80x-128}{x^2-4x+4}$
$=\dfrac{(x^2-4x+4)(x^3+4x^2+12x+32)+80x-128}{x^2-4x+4}$
$=x^3+4x^2+12x+32+\dfrac{80x-128}{x^2-4x+4}$
We will write the partial fraction decomposition for the fraction $\dfrac{80x-128}{x^2-4x+4}$. Factor the denominator and write the partial fraction decomposition:
Factor the denominator:
$\dfrac{80x+128}{x^2-4x+4}=\dfrac{80x-128}{(x-2)^2}=\dfrac{A}{x-2}+\dfrac{B}{(x-2)^2}$
Identify the coefficients:
$80x+128=A(x-2)+B$
$80x+128=Ax-2A+B$
$80x+128=Ax+(-2A+B)$
$\begin{cases}
A=80\\
-2A+B=128
\end{cases}$
$-2(80)+B=128$
$-160+B=128$
$B=288$
The partial fraction decomposition is:
$\dfrac{x+2}{(x-1)(x+1)}=\dfrac{\dfrac{3}{2}}{x-1}-\dfrac{\dfrac{1}{2}}{x+1}$
$\dfrac{x+2}{(x-1)(x+1)}=\dfrac{3}{2(x-1)}-\dfrac{1}{2(x+1)}$
The final result is:
$\dfrac{x^5}{x^2-4x+4}=x^3+4x^2+12x+32+\dfrac{80}{x-2}+\dfrac{288}{(x-2)^2}$
