Question

Divide \[y^4 - 3 y^3 + \frac{1}{2} y^2 by 3y\]

Answer 1

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