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

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hartnnBest ResponseYou've already chosen the best response.0
this is your question or you just testing other's knowledge again ?
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
Sorry if that is i nappropriate!
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
I'll close the question if this is inappropriate.
 6 months ago

hartnnBest 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...
 6 months ago

hartnnBest ResponseYou've already chosen the best response.0
more important is that you do not repeat this again
 6 months ago

nincompoopBest ResponseYou've already chosen the best response.0
alt+F4 command works better
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
I won't do that again if it is inapp...
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
Not this equation, but I can certainly do it.
 6 months ago

hartnnBest ResponseYou've already chosen the best response.0
see ? you just wasted 2 minutes of dan. thats the harm i am talking about...
 6 months ago

hartnnBest ResponseYou've already chosen the best response.0
he could have easily helped a needy fellow
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
Sorry guys! Ignore my math questions if other than calc!
 6 months ago

dan815Best ResponseYou've already chosen the best response.1
sign error it shud be 5
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
What I am getting is that dw:1381870420911:dw
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
dan815, I don't help, really ignore me!
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
b/c remainder is 4, that's how....
 6 months ago

myininayaBest 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
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
I knew how to do it from the beginning...closed the question... Sorry!
 6 months ago

myininayaBest 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
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
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.
 6 months ago

myininayaBest 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
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
Yes. U see what i did..... I know it 100% don't worry.
 6 months ago

myininayaBest ResponseYou've already chosen the best response.2
Ok. I'm not worried. I'm only responding here because I wanted to.
 6 months ago

myininayaBest ResponseYou've already chosen the best response.2
And also just in case someone else can learn from it.
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
let other ppl... they do
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
know this, since they came to help..... But true, they were making it much more complicated than what it is.
 6 months ago

SolomonZelmanBest ResponseYou've already chosen the best response.0
(add to my prev comment)
 6 months ago

dan815Best 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
 6 months ago

myininayaBest ResponseYou've already chosen the best response.2
lol. I was just trying to help....
 6 months ago

dan815Best ResponseYou've already chosen the best response.1
well it was helpful! i didnt think of just subbing in 2
 6 months ago

myininayaBest ResponseYou've already chosen the best response.2
Your way was cute too.
 6 months ago

dan815Best 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
 6 months ago

myininayaBest ResponseYou've already chosen the best response.2
I actually think it is called the remainder theorem.
 6 months ago

dan815Best ResponseYou've already chosen the best response.1
oh ive heard of that before, thanks!
 6 months ago

myininayaBest ResponseYou've already chosen the best response.2
http://en.wikipedia.org/wiki/Polynomial_remainder_theorem
 6 months ago
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