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anonymous
 one year ago
The smallest integer that can be added to 2m^3 − m + m^2 + 1 to make it completely divisible by m + 1 is
anonymous
 one year ago
The smallest integer that can be added to 2m^3 − m + m^2 + 1 to make it completely divisible by m + 1 is

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you comfortable with synthetic divison or long division?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it doesnt matter haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, making sure. So let's divide 2m^3 +m^2 m +1 by m+1 and see what we get. Ill use synthetic. dw:1433724888539:dw So this division gives us: \[2m^{2} + 3m 4 + \frac{ 5 }{ m+1 }\] So what this means is as long as 5/(m+1) reduces, we're good. Since we're allowed to use any integer, we can make m = 6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so is 6 the answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would say so. m+1 will only be divide into your polynomial if m+1 divides 5, since that is the remainder term we have. So either m +1 = 5 or m+1 = 5 would work, the smalles of them being m + 1 = =5, m = 6

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0I think they are asking what is the smallest integer that could be added to 2m^3 +m^2 m +1 to make it so when divided by m+1 you have no remainder. Since we have a remainder of 5, if were to add 5 + m +1 = m4 to the cubic, we would have no remainder and then division by m+1 would be possible (i.e. no remainder). This would give $$ 2m^3 +m^2 m +1 + m 4 = 2m^3+m^23 $$ When we divide this by m+1 we get http://www.wolframalpha.com/input/?i=%282m%5E3%2Bm%5E23%29%2F%28m%2B1%29 So adding m4 to the cubic makes it divisible by m+1.
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