Answer
$= \frac{3x}{4} - 2 + \frac{1}{4x}$
Work Step by Step
$\frac{3x^{4}-8x^{3}+x^{2}}{4x^{3}}$
$= \frac{x^{2}(3x^{2}-8x+1)}{x^{2}(4x)}$
$= \frac{3x^{2}-8x+1}{4x}$
$= \frac{3x^{2}}{4x} - \frac{8x}{4x} + \frac{1}{4x}$
$= \frac{3x}{4} - 2 + \frac{1}{4x}$
