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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Solve for k.
 9 months 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!
 9 months 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
 9 months 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.
 9 months 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
 9 months 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.
 9 months 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.
 9 months ago

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

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

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

SolomonZelman Group TitleBest ResponseYou've already chosen the best response.1
WHATEVER.
 9 months 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
 9 months ago

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

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

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

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

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

myininaya Group TitleBest ResponseYou've already chosen the best response.2
Theorems.
 9 months ago

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

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

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