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)

Tutors, please sign in to answer this question.

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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