Algebra
posted by Aaron .
Divide
(42b^3 + 23b^2 + 38b + 49)/ (6b + 5)

Algebra 
David Q/R
Treat it like a long division. First, work out how many times 6b goes into 42b^3, which is 7b^2, and that will be the first term of your answer. Now multiply (6b+5) by 7b^2, and you'll get 42b^3 + 35b^2. Write that underneath the original expression and subtract it from it: you'll get (12b^2 + 38b + 49).
Next, you need to work out how many times (6b+5) goes into that. To find that, ask how many times 6b goes into 12b^2, which is 2b, and that will be the next term in your answer. Multiply (6b+5) by 2b, and you'll get (12b^2  10b). Subtract that from (12b^2 + 38b + 49), and you'll get (48b + 49). But now you've got a problem, because(6b+5) doesn't divide that exactly: the closest you can get is 8, which would make your answer (7b^2  2b + 8). But that would only work if your final term were 40 instead of 49. Are you sure you haven't copied it down wrong?
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