#### notes

Let us consider the division of the trinomial `4y^3 + 5y^2 + 6y` by the monomial 2y.

`4y^3 + 5y^2 + 6y = (2 × 2 × y × y × y) + (5 × y × y) + (2 × 3 × y)`

we find that 2 × y is common in each term. Therefore, separating 2 × y from each term. We get

`4y^3 + 5y^2 + 6y =2 × y × (2 × y × y) + 2 × y × (5/2 xx y) + 2 xx y xx 3`

`= 2y (2y^2) + 2y (5/(2 y)) + 2y(3)`

`= 2y (2y^2) + 2y (5/(2y)) 2y(3)`

`= 2y (2y^2 + 5/(2 y) +3)` ...(the common factor 2y is shown separately).

Alternatively, we could divide each term of the trinomial by the monomial using the cancellation method.

`(4y^3 + 5y^2 + 6y) ÷ 2y = (4y^3 + 5y^2 + 6y)/(2y)`

`= (4y^3)/(2y) + (5y^2)/(2y) + (6y)/(2y)`

` = 2y^2 + 5/(2y) + 3`