SOLUTION: I am lost on this problem: (35b^(3)+25b^(2)+30b+47)÷(5b+5) I started with the first part (35b^(3)+25b^(2)+30b+47)and I tried to divide each part by the second part (5b) (

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Question 387337: I am lost on this problem: (35b^(3)+25b^(2)+30b+47)÷(5b+5)
I started with the first part
(35b^(3)+25b^(2)+30b+47)and I tried to divide each part by the second part
(5b) (5b) (5b) (5) but none of my answers were the same as the possible answers.


Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
I get an answer of 7b^2 - 2b + 8
first you divide 35b^3 by 5b to get 7b^2
then you multiply (5b+5) by 7b^2 to get (35b^3 + 35b^2)
then you subtract (35b^3 + 35b^2) from (35b^3 + 25b^2) to get -10b^2
then you bring down the 30b to get (-10b^2 + 30b)
then you divide -10b^2 by 5b to get -2b
then you multiply (5b+5) by -2b to get (-10b^2 - 10b)
then you subtract (-10b^2 - 10b) from (-10b^2 + 30b) to get 40b
then you bring down the 47 to get (40b + 47)
then you divide 40b by 5b to get 8
then you multiply (5b+5) by 8 to get (40b + 40)
then you subtract (40b + 40) from (40b + 47) to get 7
that's your remainder.

the answer is 7b^2 - 2b + 8 with a remainder of 7

on paper, it looks like what is shown below.

***** picture not found *****

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