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F(x) question

  1. May 25, 2004 #1
    If i have f(x) = 2x^3 + 7x -3

    And when f(x) is divided by (2x-k) I have -8

    How do i find values for k?
     
  2. jcsd
  3. May 25, 2004 #2
    I might be misunderstanding your question, but are you saying that

    [tex]\frac{2x^3+7x-3}{2x-k}=-8[/tex]

    and you wish to solve for k as a function of x?
     
  4. May 25, 2004 #3

    Gokul43201

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    How do you actually divide 2 polynomials ? Do the long division ...you'll find the remaider to be R = f(k), where f is a cubic in k. Equate this to -8 and you have the 3 values of k.
     
  5. May 25, 2004 #4

    matt grime

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    Yes and no there's no need to solve as a function of x: he's saying that if we write

    2x^3+7x-3 = (2x-k)(x^2+ax+b)-8

    what is k?

    It's a division algorithm for polynomials question; by comparing coeffs on each side you ought to be able to solve a set of simulatneous equations to find k (and a and b)
     
  6. May 25, 2004 #5

    Gokul43201

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    I'm assuming you mean the remainder is -8. It's not clear, and evidently, TALewis has interpreted differently.

    And Matt's way is much easier.
     
    Last edited: May 25, 2004
  7. May 25, 2004 #6
    Ah, if you mean the remainder is -8, that makes much more sense.
     
  8. May 25, 2004 #7
    yes, what i mean is that when, 2x^3+7x-3/(2x-k) This equals some polynomial funtion with a remainder of -8. How do i work out what the polynomial is?????????
     
    Last edited: May 25, 2004
  9. May 25, 2004 #8

    Gokul43201

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    Look at matt's post again.

    Expand the RHS and equate coefficients of the different powers of x, between the LHS and RHS. For instance, "Is there an x^2 term in the LHS ?...No !" So you equate the coefficient in the RHS to zero...and so on. Solving from these equations gives you k.
     
  10. May 26, 2004 #9
    Yes youth. Cheers !
     
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