Answer
$y^{2}-3y+\dfrac{5}{4}$
Work Step by Step
Dividing each of the terms of the numerator by the denominator, the given expression, $
\dfrac{4y^3-12y^2+5y}{4y}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{4y^3}{4y}-\dfrac{12y^2}{4y}+\dfrac{5y}{4y}
.\end{array}
Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to
\begin{array}{l}\require{cancel}
y^{3-1}-3y^{2-1}+\dfrac{5\cancel{y}}{4\cancel{y}}
\\\\=
y^{2}-3y^{1}+\dfrac{5}{4}
\\\\=
y^{2}-3y+\dfrac{5}{4}
.\end{array}