Answer
$\frac{2x^3-10x^2+16x-2}{(x-2)^2}$
Work Step by Step
$\left(\frac{f}{g}\right)'$ = $\left(\frac{f'g-fg'}{g^2}\right)$
$\left(\frac{x^3-4x^2+x}{x-2}\right)'$ = $\frac{(3x^2-8x+1)(x-2)-(x^3-4x^2+x)(1)}{(x-2)^2}$ = $\frac{2x^3-10x^2+16x-2}{(x-2)^2}$
