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Homework Help: Yet another dumb question (zeros of a cubic function)

  1. Dec 3, 2007 #1
    f(x)= x^3-2x^2-11x+52 ----> Find all of the zeros of the function

    I just need some advice/direction on how to start solving this. It is not in the textbook and my brain is just kinda farting when I look at it. I think i need to get it in an (x-a)(bx^2+cx+d) form or something, but I dont know.
  2. jcsd
  3. Dec 3, 2007 #2


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    I'd try and guess a factor to start with, looking at the factors of 52; so +/-2, +/-4, etc.. then divide the polynomial by (x-a) where a is a factor.
  4. Dec 3, 2007 #3
    Here's another idea you can try:

    Just test out a few x values and see if you can find two values (say a and b) such that f(a) is, say, positive and f(b) is negative (or vice versa). Then, you know that since f(a) and f(b) are of opposite signs, somewhere in between is a value c in between a and b such that f(c) = 0.

    For example, take g(x) = x^2 - 4 (this should be easy enough). If I test out:

    g(1) = 1-4 = -3 < 0
    g(7) = 49-4 = 45 > 0

    Then I know that between 1 and 7 there is a zero (because the y-values go from -3, which is negative, to 45, which is positive, and hence at some point the y-values had to be 0). You can narrow your root even more by adjusting the interval (instead of 7, try smaller values so that the y-values are even closer to 0).

    This concept is an application of the intermediate value theorem.
  5. Jan 6, 2008 #4


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    I would use a graphic plotter if you have one and plot the cubic. You then see that x = -4 looks like a solution and you soon check that it is exactly. Most graphic plotters have an root-finding button. (Otherwise you arrive at this by the calculations of integer x already suggested.) That is (x + 4) is a factor of the cubic. So divide by this and you find the other factors (x^2 - 6x + 13). This has no real solutions as you will have already seen from the cubic plot. You can get the complex roots solving the quadratic; they are 3 + 2i, 3 - 2i.
  6. Jan 6, 2008 #5


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    I think you kinda just gave away the answer... usually we try an learn 'em a little, not just give them the answer... but to each his own, as they say.
  7. Jan 6, 2008 #6


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    I guess I did, but maybe he has other problems like that. I guess I had doubts about whether there really is any significant learnin in a problem like this that IMHO merited only cheating!:biggrin: - but on second thoughts maybe there is some, but we wouldn't want him to be looking for ever for real roots that are not there.:smile:
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