Answer
$\frac{x^3-2x^2-4x+3}{x^4}=\frac{1}{x}+\frac{-2}{x^2}+\frac{-4}{x^3}+\frac{3}{x^4}$
Work Step by Step
$\frac{x^3-2x^2-4x+3}{x^4}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}$,
$Ax^3+Bx^2+Cx+D=x^3-2x^2-4x+3$,
$A=1, B=-2, C=-4, D=3$
thus,
$\frac{x^3-2x^2-4x+3}{x^4}=\frac{1}{x}+\frac{-2}{x^2}+\frac{-4}{x^3}+\frac{3}{x^4}$