Question
Divide $\text{y}^4-3\text{y}^4+\frac{1}{2}\text{y}^2\text{ by }3\text{y}.$

Answer

$\frac{\text{y}^4-3\text{y}^3+\frac{1}{2}\text{y}^2}{3\text{y}}$
$=\frac{\text{y}^4}{3\text{y}}-\frac{3\text{y}^3}{3\text{y}}+\frac{\frac{1}{2}\text{y}^2}{3\text{y}}$
$=\frac{1}{3}\text{y}^{(4-1)}-\text{y}^{(3-1)}+\frac{1}{6}\text{y}^{(2-1)}$
$=\frac{1}{3}\text{y}^3-\text{y}^2+\frac{1}{6}\text{y}$

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