Question:
Divide $y^{4}-3 y^{3}+\frac{1}{2} y^{2}$ by $3 y$.
Solution:
$\frac{y^{4}-3 y^{3}+\frac{1}{2} y^{2}}{3 y}$
$=\frac{y^{4}}{3 y}-\frac{3 y^{3}}{3 y}+\frac{\frac{1}{2} y^{2}}{3 y}$
$=\frac{1}{3} y^{(4-1)}-y^{(3-1)}+\frac{1}{6} y^{(2-1)}$
$=\frac{1}{3} y^{3}-y^{2}+\frac{1}{6} y$
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