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SolomonZelman
Group Title
x^3+Kx^2+12x8, is divisible by (x2)
so that the remainder is 4 what is K?
 one year ago
 one year ago
SolomonZelman Group Title
x^3+Kx^2+12x8, is divisible by (x2) so that the remainder is 4 what is K?
 one year ago
 one year ago

This Question is Closed

hartnn Group TitleBest ResponseYou've already chosen the best response.0
this is your question or you just testing other's knowledge again ?
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
Sorry if that is i nappropriate!
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
I'll close the question if this is inappropriate.
 one year ago

nincompoop Group TitleBest ResponseYou've already chosen the best response.0
very inappropriate
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
lol, how many times i need to tell you that its not inappropriate, but its not required! and i explained what harm can 'not required' do...
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
Close?
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.1
dw:1381783626778:dw
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
OK I'm closing?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
more important is that you do not repeat this again
 one year ago

nincompoop Group TitleBest ResponseYou've already chosen the best response.0
alt+F4 command works better
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
I won't do that again if it is inapp...
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.1
so u know that
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
yep!
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
Not this equation, but I can certainly do it.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
see ? you just wasted 2 minutes of dan. thats the harm i am talking about...
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
he could have easily helped a needy fellow
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
OK I'll close.
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
just did
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
Sorry guys! Ignore my math questions if other than calc!
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.1
dw:1381783979271:dw
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.1
sign error it shud be 5
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
What I am getting is that dw:1381870420911:dw
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
See why?
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
dan815, I don't help, really ignore me!
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
b/c remainder is 4, that's how....
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Well we could have done this much easier way... let f be the expression you were fiven f(2)=4 solve for k .... \[2^3+k(2)^2+12(2)8=8+4k+248=4k+24\] 4k+24=4
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Solve for k.
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
I knew how to do it from the beginning...closed the question... Sorry!
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
I just wanted to make sure you guys know it wasn't as complicated as you were making it. :P
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
For sure not! Another way to do it is synthetic division... 8+2(16+2)k=4 the remainder I get it don't worry.
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Yep. That is a way to do it but I think it easier to do if given (poly of higher degree than 0)/(xc), then all we need to do is that poly evaluated at c to find the remainder. f(c)=Remainder
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
Yes. U see what i did..... I know it 100% don't worry.
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Ok. I'm not worried. I'm only responding here because I wanted to.
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
And also just in case someone else can learn from it.
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
let other ppl... they do
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
know this, since they came to help..... But true, they were making it much more complicated than what it is.
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
(add to my prev comment)
 one year ago

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
WHATEVER.
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.1
calm down guys none of us like PHD students here . we're all noobs believe it or not
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
lol. I was just trying to help....
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
on a lost cause :P
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Looks like it.
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.1
well it was helpful! i didnt think of just subbing in 2
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Your way was cute too.
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.1
how did u know that subbing in that value gives you the remainder though? like the way if its 0 then it is a good factor
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Theorems.
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
I actually think it is called the remainder theorem.
 one year ago

dan815 Group TitleBest ResponseYou've already chosen the best response.1
oh ive heard of that before, thanks!
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.2
http://en.wikipedia.org/wiki/Polynomial_remainder_theorem
 one year ago
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