Suppose we want to divide

4y
^{
3
}
+ 5y
^{
2
}
+ 6y ÷ 2y

We first factorise both of them separately

4y
^{
3
}
+ 5y
^{
2
}
+ 6y = 4 × y
^{
3
}
+ 5 × y
^{
2
}
+ 6 × y

= 2 × 2 × y × y × y + 5 × y × y + 2 × 3 × y

2y = 2 × y

Now, Dividing

(4y
^{
3
}
+ 5y
^{
2
}
+ 6y)/2y = (2 × 2 × y × y × y + 5 × y × y + 2 × 3 × y)/(2 × y)

= (2 × 2 × y × y × y)/(2 × y) + (5 × y × y )/(2 × y) + (2 × 3 × y)/(2 × y)

= 2 × y × y + 5/2 × y + 3

= 2y
^{
2
}
+ 5/2 y + 3

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