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SolomonZelman

  • one year ago

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

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  1. hartnn
    • one year ago
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    this is your question or you just testing other's knowledge again ?

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

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

  4. nincompoop
    • one year ago
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    very inappropriate

  5. hartnn
    • one year ago
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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...

  6. SolomonZelman
    • one year ago
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    Close?

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

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

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

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

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

  12. dan815
    • one year ago
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    so u know that

  13. SolomonZelman
    • one year ago
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    yep!

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

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

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

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

  18. dan815
    • one year ago
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    lol

  19. SolomonZelman
    • one year ago
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    just did

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

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

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

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

  24. SolomonZelman
    • one year ago
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    See why?

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

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

  27. myininaya
    • one year ago
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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

  28. myininaya
    • one year ago
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    Solve for k.

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

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

  31. SolomonZelman
    • one year ago
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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.

  32. myininaya
    • one year ago
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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

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

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

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

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

  37. myininaya
    • one year ago
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    What?

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

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

  40. SolomonZelman
    • one year ago
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    WHATEVER.

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

  42. dan815
    • one year ago
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    lol

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

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

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

  46. myininaya
    • one year ago
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    Looks like it.

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

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

  49. dan815
    • one year ago
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    :)

  50. dan815
    • one year ago
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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

  51. myininaya
    • one year ago
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    Theorems.

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

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

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

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