Answer
$y^4-\dfrac{1}{3}y+4$
Work Step by Step
Divide each term of the trinomial by the divisor to obtain:
$\require{cancel}
=\dfrac{3y^5}{3y} - \dfrac{y^2}{3y}+\dfrac{12y}{3y}$
Cancel common factors and divide the variables using the quotient rule for exponents ($\frac{a^m}{a^n}=a^{m-n}$) to obtain:
$\require{cancel}
\\=\dfrac{\cancel{3}y^5}{\cancel{3}y} - \dfrac{y^2}{3y}+\dfrac{\cancel{12}4\cancel{y}}{\cancel{3}\cancel{y}}
\\=\dfrac{y^5}{y}-\dfrac{y^2}{3y}+4
\\=y^{5-1}-\dfrac{1}{3}y^{2-1}+4
\\=y^4-\dfrac{1}{3}y+4$
