SOLUTION: find the quotient for (14y+8y^2+y^3+12)÷(6+y)

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Question 311190: find the quotient for (14y+8y^2+y^3+12)÷(6+y)
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
%28y%5E3%2B8y%5E2%2B14y%2B12%29%2F%28y%2B6%29
First factor y%5E2
y2%28y%2B6%29=y%5E3%2B6y%5E2
Subtracted from the original polynomial gives a remainder of,
%28y%5E3%2B8y%5E2%2B14y%2B12%29-%28y%5E3%2B6y%5E2%29=2y%5E2%2B14y%2B12
Second factor 2y
2y%28y%2B6%29=2y%5E2%2B12y
Subtracted from the previous remainder gives a new remainder of,
%282y%5E2%2B14y%2B12%29-%282y%5E2%2B12y%29=2y%2B12
Third factor: 2
2%28y%2B6%29=2y%2B12
Subtracted from the previous remainder gives a remainder of 0.
2y%2B12-%282y%2B12%29=0
Gather up all of the factors.
%28y%5E3%2B8y%5E2%2B14y%2B12%29%2F%28y%2B6%29=y%5E2%2B2y%2B2