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) (
Algebra ->
Polynomials-and-rational-expressions
-> 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.
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.
