1. (12y

^{2}- 39y+15+12y^{3 })/(2y+5)4. (w

^{3}- 15w-11)/(w-4)-
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1. (12y^{2} - 39y+15+12y^{3 })/(2y+5)

4. (w^{3} - 15w-11)/(w-4)

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Nkeiruka, we do this just like we do long division with whole numbers. Let's see what we can come up with.

First, we have to put things in order: 12y^3 + 12y^2 - 39y +15

How many times does 2y go into 12y^3? 6y^2, right? 6y^2(2y+5)= 12y^3+30y^2.

12y^3 - 12y^2

- 12y^3 +30y^2

______________

-18y^2

Bring down the -39y, so you have -18y^2 - 39y. How many times does 2y go into -18y? -9y, right? -9y(2y+5)= -18y^2-45y

-18y^2 - 39y

- -18y^2 - 45y

______________

6y

Bring down the 15 to make 6y+15

How many times does 2y go into 6y? 3, so 3(2y+5)= 6y+15

6y + 15

- 6y + 15

__________

0

You final answer is 6y^2 - 9y + 3.

Check your answer and do the same for the second problem.

Nkeiruka,

How are you setting up your solution to these questions? Maybe we can help you where you're stuck.

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