SolomonZelman
x^3+Kx^2+12x-8, is divisible by (x-2)
so that the remainder is 4 what is K?
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hartnn
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this is your question or you just testing other's knowledge again ?
SolomonZelman
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Sorry if that is i nappropriate!
SolomonZelman
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I'll close the question if this is inappropriate.
nincompoop
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very inappropriate
hartnn
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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...
SolomonZelman
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Close?
dan815
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|dw:1381783626778:dw|
SolomonZelman
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OK I'm closing?
hartnn
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more important is that you do not repeat this again
nincompoop
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alt+F4 command works better
SolomonZelman
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I won't do that again if it is inapp...
dan815
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so u know that
SolomonZelman
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yep!
SolomonZelman
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Not this equation, but I can certainly do it.
hartnn
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see ? you just wasted 2 minutes of dan.
thats the harm i am talking about...
hartnn
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he could have easily helped a needy fellow
SolomonZelman
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OK I'll close.
dan815
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lol
SolomonZelman
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just did
SolomonZelman
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Sorry guys! Ignore my math questions if other than calc!
dan815
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|dw:1381783979271:dw|
dan815
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sign error it shud be -5
SolomonZelman
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What I am getting is that
|dw:1381870420911:dw|
SolomonZelman
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See why?
SolomonZelman
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dan815, I don't help, really ignore me!
SolomonZelman
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b/c remainder is 4, that's how....
myininaya
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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
myininaya
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Solve for k.
SolomonZelman
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I knew how to do it from the beginning...closed the question...
Sorry!
myininaya
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I just wanted to make sure you guys know it wasn't as complicated as you were making it. :P
SolomonZelman
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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.
myininaya
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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
SolomonZelman
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Yes.
U see what i did.....
I know it 100% don't worry.
myininaya
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Ok. I'm not worried. I'm only responding here because I wanted to.
myininaya
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And also just in case someone else can learn from it.
SolomonZelman
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let other ppl... they do
myininaya
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What?
SolomonZelman
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know this, since they came to help..... But true, they were making it much more complicated than what it is.
SolomonZelman
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(add to my prev comment)
SolomonZelman
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WHATEVER.
myininaya
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Ok....
dan815
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lol
dan815
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calm down guys none of us like PHD students here -.- we're all noobs believe it or not
myininaya
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lol. I was just trying to help....
hartnn
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on a lost cause :P
myininaya
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Looks like it.
dan815
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well it was helpful! i didnt think of just subbing in 2
myininaya
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Your way was cute too.
dan815
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:)
dan815
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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
myininaya
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Theorems.
myininaya
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I actually think it is called the remainder theorem.
dan815
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oh ive heard of that before, thanks!