Question 236119
I am supposed to divide the polynomial: 
(42b^3+23b^2+32b+38)/(6b+5) 
The answer I get is 7b^2-2b+7 with a remainder of 3- would that be written as 7b^2-2b+7+3/6b+5, or would it be written as 7b^2-2b+7-3/6b+5?
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If you were dividing 7 by 3 you would get 2 with a remainder of 1
You could write that as 2 + 1/3
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Your Problem:
(numerator)/(denominator) = quotient + (remainder/denominator)
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(42b^3....)(6b+5) = (7b^2...) + 3/(6b+5)
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Cheers,
Stan H.