anonymous
  • anonymous
The smallest integer that can be added to -2m^3 − m + m^2 + 1 to make it completely divisible by m + 1 is
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Are you comfortable with synthetic divison or long division?
anonymous
  • anonymous
it doesnt matter haha
anonymous
  • anonymous
Okay, 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

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anonymous
  • anonymous
so is -6 the answer?
anonymous
  • anonymous
I 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
anonymous
  • anonymous
oh okay haha thanks
anonymous
  • anonymous
No problem :)
ybarrap
  • ybarrap
I 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 = m-4 to the cubic, we would have no remainder and then division by m+1 would be possible (i.e. no remainder). This would give $$ -2m^3 +m^2 -m +1 + m -4 = -2m^3+m^2-3 $$ When we divide this by m+1 we get http://www.wolframalpha.com/input/?i=%28-2m%5E3%2Bm%5E2-3%29%2F%28m%2B1%29 So adding m-4 to the cubic makes it divisible by m+1.

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