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Find the remainder when 1/f(x) is divided by x+1

  1. May 25, 2005 #1
    It is given that f(x)=8x^3+4x-3. The question is: Find the remainder when 1/f(x) is divided by x+1. My text book says the remainder does not exist!?! I just cant solve it.
    Thanks in advance for any help.
  2. jcsd
  3. May 25, 2005 #2


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    Exactly what is meant by "remainder" here?

    For example, if I ask "what is the remainder when 17 is divided by 6" the answer is 5:
    6 divides into 17 twice with remainder 17-12= 5. That's because I'm working with integers.

    However, if I ask "what is the remainder when 1/17 is divided by 6", the question no longer makes sense. 1/17 is not an integer so I cannot be talking about integer operations- and there is no "remainder" in rational number division.

    In algebra, "polynomials" correspond to "integers" and "rational functions" to rational numbers. If the problem were "what is the remainder when 8x^3+ 4x- 3 is divided by x+1" it's easy to see that the answer is -15: 8x^3+ 4x- 3= (x+1)(8x^2- 8x+ 12)- 15.

    But I don't know what is meant by "remainder" in dividing the rational function 1/(8x^3+ 4x- 3) by x+1.
  4. May 25, 2005 #3
    Maybe that is exactly what hes lookin for.
  5. May 25, 2005 #4
    I think he's supposed to use polynomial long division or something called like this. Anywho, it might help you if you looked at it like so
    [tex]\frac{(1/(8x^3+4x-3))}{(x+1)} [/tex] is also [tex] \frac{1}{(x+1)(8x^3+4x-3)}[/tex].
    The way i remember it, you have to find what you would multiply the denomenator by to get the numerator(sort of useless really), but the above fraction has a definite answer with no remainder. For an example of the "long division" (x^2+3x-5)/(x+1) = x+2 - (7/(x+1))*
    *this would be the remainder.
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