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x^3+Kx^2+12x-8, is divisible by (x-2) so that the remainder is 4 what is K?

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this is your question or you just testing other's knowledge again ?
Sorry if that is i nappropriate!
I'll close the question if this is inappropriate.

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Other answers:

very inappropriate
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...
OK I'm closing?
more important is that you do not repeat this again
alt+F4 command works better
I won't do that again if it is inapp...
so u know that
Not this equation, but I can certainly do it.
see ? you just wasted 2 minutes of dan. thats the harm i am talking about...
he could have easily helped a needy fellow
OK I'll close.
just did
Sorry guys! Ignore my math questions if other than calc!
sign error it shud be -5
What I am getting is that |dw:1381870420911:dw|
See why?
dan815, I don't help, really ignore me!
b/c remainder is 4, that's how....
Well we could have done this much easier way... let f be the expression you were fiven f(2)=4 solve for k .... \[2^3+k(2)^2+12(2)-8=8+4k+24-8=4k+24\] 4k+24=4
Solve for k.
I knew how to do it from the beginning...closed the question... Sorry!
I just wanted to make sure you guys know it wasn't as complicated as you were making it. :P
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.
Yep. That is a way to do it but I think it easier to do if given (poly of higher degree than 0)/(x-c), then all we need to do is that poly evaluated at c to find the remainder. f(c)=Remainder
Yes. U see what i did..... I know it 100% don't worry.
Ok. I'm not worried. I'm only responding here because I wanted to.
And also just in case someone else can learn from it.
let other ppl... they do
know this, since they came to help..... But true, they were making it much more complicated than what it is.
(add to my prev comment)
calm down guys none of us like PHD students here -.- we're all noobs believe it or not
lol. I was just trying to help....
on a lost cause :P
Looks like it.
well it was helpful! i didnt think of just subbing in 2
Your way was cute too.
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
I actually think it is called the remainder theorem.
oh ive heard of that before, thanks!

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