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Homework Help: Zeroes of a polynomial

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data

    f(x) = x^3 + 4x^2 − 4x − 16.
    a) How many possible rational zeros does f(x) have?
    1. None
    2. Two
    3. Ten
    4. Three
    5. Five

    3. The attempt at a solution

    My first response was answer 4. - Three zeroes. I
    a) solved for the zeroes on paper
    b) graphed it on a graphing calculator
    c) realized that a third degree polynomial can have three zeroes
    d) answered the next question, asking what the zeroes were, correctly.

    Apparently I'm misunderstanding the question though, because answer 4 is not correct. So just what is the question asking? I thought it was simply "how many zeroes does it have", but I guess I'm missing something.
    Last edited: Nov 2, 2009
  2. jcsd
  3. Nov 2, 2009 #2
    The question asks: How many possible rational zeros does f(x) have? I think it is a bad question though, because the polynomial has a possibly of having 3 zeros, and if I'm not mistaken, all of these zeros could potentially be rational.
  4. Nov 2, 2009 #3
    All of the zeroes are rational - I'm going off memory, but I believe they were -4, -2, and 2. Regardless, they were all integers. That's why I'm so confused.
  5. Nov 2, 2009 #4
    Then it appears that the answer is incorrect.
  6. Nov 2, 2009 #5
    It is possible that the question is asking how many different values are possible for a rational root from the rational roots theorem. I agree that it is a poorly worded question.
  7. Nov 2, 2009 #6
    So the answer is 5...correct?
  8. Nov 2, 2009 #7
    Actually, I remembered that rational roots allows both positive and negative roots, so I would say 10... but who knows how this question is to be interpreted. Five is the closest to any sort of right answer if three isn't correct.
  9. Nov 2, 2009 #8


    Staff: Mentor

    The question might be asking how many rational candidates are there for zeroes in this cubic. If that's the correct interpretation, then the answer is ten. The candidates are [itex]\pm[/itex] 1, [itex]\pm[/itex] 2, [itex]\pm[/itex] 4, [itex]\pm[/itex] 8, and [itex]\pm[/itex] 16.
  10. Nov 2, 2009 #9
    Thought about it some more and remembered it's plus or minus, came on here and read what I was thinking.

    Thanks guys
  11. Nov 3, 2009 #10


    User Avatar
    Science Advisor

    As n!kofeyn said, that is a badly worded question. Obviously, a cubic equation cannot have more than 3 zeros and so not more than three possible rational zeros.

    However, there there are 10 rational numbers that, by the rational number theorem, are possible zeros.
  12. Nov 3, 2009 #11
    You can always substitute x = y + p and then obtain other possible rational solutions. The intersection of all these sets of solutions is the correct answer, so I think the correct answer is 3 if there are 3 rational solutions. In fact, you can save a lot of time by shifting your variable. E.g., if you see if x = 1 is a solution you find that:

    f(1) = -15

    So, x = 1 is not a zero. But this result is nevertheless useful, becaus it is clear that the polynomial g(y) defined as:

    g(y) = f(1+y)

    has 1 as the coefficient of y^3 and the constant term is

    g(0) = f(1) = -15

    So, if we apply the Rational Roots Theorem to g(y), we find that the zeroes of g(y) can be:

    y = ±1, ±3, ±5,±15

    Since g(y) = f(1+y), this means that we need to add 1 to these numbers to obtain the possible rational zeroes of f(x):

    x = 0, 2, -2, 4, -4, 6, -14, 16

    If you strike out those candidates that are not on the original list, you are left with:

    x = 2, -2, 4, -4, 16

    If you play the same game with x = -1, you find:

    f(-1) = -9

    therefore putting y = -1 + x gives the candidates:

    y = ±1, ±3, ±9


    x = -2, 0, -4, 2, -10, 8

    Striking out those candidates that are not on the previous list, leaves us with:

    x = -2, -4, 2

    Which are the three correct solutions.
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