Answer
$\dfrac{3x}{2}-1+\dfrac{1}{2x}$
Work Step by Step
Given: $(6x^3-4x^2+2x) \times \dfrac{1}{4x^2}$
$(6x^3-4x^2+2x) \times \dfrac{1}{4x^2}=\dfrac{(6x^3-4x^2+2x)}{4x^2}$
$=\dfrac{2x(3x^2-2x+1)}{4x^2}$
$=\dfrac{(3x^2-2x+1)}{2x}$
$=\dfrac{3x}{2}-1+\dfrac{1}{2x}$
