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2m-1 divided by 4m^3 - 8m^2 + 4m + 6
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What would \(2m\) have to be multiplied by to get \(4m^3\)? That is your first term. (Also, are you sure it isn't \(4m^3-8m^2+4m+6\) divided by \(2m-1\)?)
[2m ^{2}-5m….\]
I think the best way to approach this question is to do long division. I just tried it and got a remander of m+6
yeah its a long division problem;; can you show me how you did it?
Step one is to answer the question: "What would \(2m\) have to be multiplied by to get \(4m^3\)?"
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i've already gotten as far as 2m^2 - 5m
|dw:1415234366978:dw| Okay, so how about 2m into 9m?
Oops 2m-1
We are looking at 2m-1 into -m+6
yes i know I got -m - \[\frac{ 1 }{ 2 }\]
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