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SolomonZelman Group Title

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

  • one year ago
  • one year ago

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  1. hartnn Group Title
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    this is your question or you just testing other's knowledge again ?

    • one year ago
  2. SolomonZelman Group Title
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    Sorry if that is i nappropriate!

    • one year ago
  3. SolomonZelman Group Title
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    I'll close the question if this is inappropriate.

    • one year ago
  4. nincompoop Group Title
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    very inappropriate

    • one year ago
  5. hartnn Group Title
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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...

    • one year ago
  6. SolomonZelman Group Title
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    Close?

    • one year ago
  7. dan815 Group Title
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    |dw:1381783626778:dw|

    • one year ago
  8. SolomonZelman Group Title
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    OK I'm closing?

    • one year ago
  9. hartnn Group Title
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    more important is that you do not repeat this again

    • one year ago
  10. nincompoop Group Title
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    alt+F4 command works better

    • one year ago
  11. SolomonZelman Group Title
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    I won't do that again if it is inapp...

    • one year ago
  12. dan815 Group Title
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    so u know that

    • one year ago
  13. SolomonZelman Group Title
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    yep!

    • one year ago
  14. SolomonZelman Group Title
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    Not this equation, but I can certainly do it.

    • one year ago
  15. hartnn Group Title
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    see ? you just wasted 2 minutes of dan. thats the harm i am talking about...

    • one year ago
  16. hartnn Group Title
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    he could have easily helped a needy fellow

    • one year ago
  17. SolomonZelman Group Title
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    OK I'll close.

    • one year ago
  18. dan815 Group Title
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    lol

    • one year ago
  19. SolomonZelman Group Title
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    just did

    • one year ago
  20. SolomonZelman Group Title
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    Sorry guys! Ignore my math questions if other than calc!

    • one year ago
  21. dan815 Group Title
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    |dw:1381783979271:dw|

    • one year ago
  22. dan815 Group Title
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    sign error it shud be -5

    • one year ago
  23. SolomonZelman Group Title
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    What I am getting is that |dw:1381870420911:dw|

    • one year ago
  24. SolomonZelman Group Title
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    See why?

    • one year ago
  25. SolomonZelman Group Title
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    dan815, I don't help, really ignore me!

    • one year ago
  26. SolomonZelman Group Title
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    b/c remainder is 4, that's how....

    • one year ago
  27. myininaya Group Title
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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

    • one year ago
  28. myininaya Group Title
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    Solve for k.

    • one year ago
  29. SolomonZelman Group Title
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    I knew how to do it from the beginning...closed the question... Sorry!

    • one year ago
  30. myininaya Group Title
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    I just wanted to make sure you guys know it wasn't as complicated as you were making it. :P

    • one year ago
  31. SolomonZelman Group Title
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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.

    • one year ago
  32. myininaya Group Title
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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

    • one year ago
  33. SolomonZelman Group Title
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    Yes. U see what i did..... I know it 100% don't worry.

    • one year ago
  34. myininaya Group Title
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    Ok. I'm not worried. I'm only responding here because I wanted to.

    • one year ago
  35. myininaya Group Title
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    And also just in case someone else can learn from it.

    • one year ago
  36. SolomonZelman Group Title
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    let other ppl... they do

    • one year ago
  37. myininaya Group Title
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    What?

    • one year ago
  38. SolomonZelman Group Title
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    know this, since they came to help..... But true, they were making it much more complicated than what it is.

    • one year ago
  39. SolomonZelman Group Title
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    (add to my prev comment)

    • one year ago
  40. SolomonZelman Group Title
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    WHATEVER.

    • one year ago
  41. myininaya Group Title
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    Ok....

    • one year ago
  42. dan815 Group Title
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    lol

    • one year ago
  43. dan815 Group Title
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    calm down guys none of us like PHD students here -.- we're all noobs believe it or not

    • one year ago
  44. myininaya Group Title
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    lol. I was just trying to help....

    • one year ago
  45. hartnn Group Title
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    on a lost cause :P

    • one year ago
  46. myininaya Group Title
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    Looks like it.

    • one year ago
  47. dan815 Group Title
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    well it was helpful! i didnt think of just subbing in 2

    • one year ago
  48. myininaya Group Title
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    Your way was cute too.

    • one year ago
  49. dan815 Group Title
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    :)

    • one year ago
  50. dan815 Group Title
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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

    • one year ago
  51. myininaya Group Title
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    Theorems.

    • one year ago
  52. myininaya Group Title
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    I actually think it is called the remainder theorem.

    • one year ago
  53. dan815 Group Title
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    oh ive heard of that before, thanks!

    • one year ago
  54. myininaya Group Title
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    http://en.wikipedia.org/wiki/Polynomial_remainder_theorem

    • one year ago
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