## Intermediate Algebra (12th Edition)

$y^{2}-3y+\dfrac{5}{4}$
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}