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Math Help - Remainder Theorm

  1. #1
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    Remainder Theorm

    When x^3+kx+5 is divided by x-2 the remainder is 3. use the remainder theorm to find the value of k.

    I have the remainder theorm, for a polynomial f(x). f(a) is the remainder when f(x) is divided by x-a.

    but im not sure how to apply the theorm the the question..thanks for your time.
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  2. #2
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by Chez_ View Post
    When x^3+kx+5 is divided by x-2 the remainder is 3. use the remainder theorm to find the value of k.

    I have the remainder theorm, for a polynomial f(x). f(a) is the remainder when f(x) is divided by x-a.

    but im not sure how to apply the theorm the the question..thanks for your time.
    By the remainder theorem, f(2) = 3, that is:

    (2)^3+k(2)+5=3

    Can you finished it off?
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  3. #3
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    theorm

    well i multiplied it out then solved it... not sure if its right tho

    8+2k+5-3=0
    8+5-3=k
    _____
    2

    = 5/2=k
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  4. #4
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    Quote Originally Posted by Chez_ View Post
    well i multiplied it out then solved it... not sure if its right tho

    8+2k+5-3=0
    8+5-3=k
    _____
    2

    = 5/2=k
    2^3+2k+5=3
    8+5-3 = -2k
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  5. #5
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    so, if its 8+5-3=-2k do i divide the whole thing by 2 and then gt 5/2
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  6. #6
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    Quote Originally Posted by Chez_ View Post
    so, if its 8+5-3=-2k do i divide the whole thing by 2 and then gt 5/2
    Sorry but I don't understand where you get \frac{5}{2} from?

    8+5-3=-2k
    10 = -2k

    What is k?
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  7. #7
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    i miss read it lol i did 8+3-3 lol that wudnt b gd in exam lolk so is k 5
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  8. #8
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    Quote Originally Posted by Chez_ View Post
    i miss read it lol i did 8+3-3 lol that wudnt b gd in exam lolk so is k 5
    Correct. If you are in anyway unsure, just plug your answer back into the function and see if you get the remainder value.

    E.G:
    f(x) =  x^3-5x+5 = 3
    f(2) = 2^3 -5(2) + 5 = 3

    EDIT: Remember to watch your sign too.
    Last edited by Simplicity; December 16th 2007 at 10:55 AM.
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