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

- 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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- jamiebookeater

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- hartnn

this is your question or you just testing other's knowledge again ?

- SolomonZelman

Sorry if that is i nappropriate!

- SolomonZelman

I'll close the question if this is inappropriate.

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## More answers

- nincompoop

very inappropriate

- hartnn

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

Close?

- dan815

|dw:1381783626778:dw|

- SolomonZelman

OK I'm closing?

- hartnn

more important is that you do not repeat this again

- nincompoop

alt+F4 command works better

- SolomonZelman

I won't do that again if it is inapp...

- dan815

so u know that

- SolomonZelman

yep!

- SolomonZelman

Not this equation, but I can certainly do it.

- hartnn

see ? you just wasted 2 minutes of dan.
thats the harm i am talking about...

- hartnn

he could have easily helped a needy fellow

- SolomonZelman

OK I'll close.

- dan815

lol

- SolomonZelman

just did

- SolomonZelman

Sorry guys! Ignore my math questions if other than calc!

- dan815

|dw:1381783979271:dw|

- dan815

sign error it shud be -5

- SolomonZelman

What I am getting is that
|dw:1381870420911:dw|

- SolomonZelman

See why?

- SolomonZelman

dan815, I don't help, really ignore me!

- SolomonZelman

b/c remainder is 4, that's how....

- myininaya

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

Solve for k.

- SolomonZelman

I knew how to do it from the beginning...closed the question...
Sorry!

- myininaya

I just wanted to make sure you guys know it wasn't as complicated as you were making it. :P

- SolomonZelman

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

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

Yes.
U see what i did.....
I know it 100% don't worry.

- myininaya

Ok. I'm not worried. I'm only responding here because I wanted to.

- myininaya

And also just in case someone else can learn from it.

- SolomonZelman

let other ppl... they do

- myininaya

What?

- SolomonZelman

know this, since they came to help..... But true, they were making it much more complicated than what it is.

- SolomonZelman

(add to my prev comment)

- SolomonZelman

WHATEVER.

- myininaya

Ok....

- dan815

lol

- dan815

calm down guys none of us like PHD students here -.- we're all noobs believe it or not

- myininaya

lol. I was just trying to help....

- hartnn

on a lost cause :P

- myininaya

Looks like it.

- dan815

well it was helpful! i didnt think of just subbing in 2

- myininaya

Your way was cute too.

- dan815

:)

- dan815

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

Theorems.

- myininaya

I actually think it is called the remainder theorem.

- dan815

oh ive heard of that before, thanks!

- myininaya

http://en.wikipedia.org/wiki/Polynomial_remainder_theorem

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