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anonymous
 one year ago
I'll FAN & MEDAL just please help!!!
The smallest integer that can be added to 2m3 − m + m2 + 1 to make it completely divisible by m + 1 is
anonymous
 one year ago
I'll FAN & MEDAL just please help!!! The smallest integer that can be added to 2m3 − m + m2 + 1 to make it completely divisible by m + 1 is

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1so we want the remainder to be 0 that is when we plug in 1 (we are pluggin in 1 because we are dividing by m(1))

freckles
 one year ago
Best ResponseYou've already chosen the best response.1hey is your thingy really \[2m^3m+m^2+1\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is this where synthetic division comes in?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you could use syntheic or easier just plug in 1 into that thing and then see what you need to add to it to make it 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[2(1)^3(1)+(1)^2+1 \text{ plus what gives you 0 }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1it might help to simplify the first part first :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1after simplifying \[2(1)^3(1)+(1)^2+1?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i used my calculator

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[2m^3m+m^2+1 \\ \text{ plug in } 1 \text{ since we have that we are dividing by } m(1) \\ 2(1)^3(1)+(1)^2+1\] \[(1)^3=(1)(1)(1)=(1)(1)=1 \\ \text{ and } \\ (1)^2=(1)(1)=1 \\ \text{ so you really have } 2(1)(1)+1+1 \\ \text{ and we also know } 2(1)=2 \text{ and } (1)=+1 \\ \text{ so we have } 2+1+1+1\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.12+3 should be 5 then you need to figure out what you can add to 5 that will give you 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.1yes so the answer should be 5 that is \[2m^3m+m^2+1+(5) \text{ divided by } m+1 \text{ should give a remainder of } 0\] it having a remainder of 0 means that one thing is divisible by (m+1)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=what+is+the+remainder+of+%282m%5E3m%2Bm%5E2%2B15%29%2F%28m%2B1%29 and as you see here the remainder is 0 which is just what we wanted

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you, i understand it now
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