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Math Help - Function: State the possible number zeros.

  1. #1
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    Question Function: State the possible number zeros.

    State the possible number of positive real zeros, negative real zeros, and imaginary zeros of each function.

    p(x) = x^5 - 6x^4 - 3x^3 + 7x^2 - 8x + 1

    h(x) = 4x^3 - 6x^2 + 8x - 5

    f(x) = x^3 - 6x^2 + 1

    I would really love to find out how to solve these problems step by step.
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  2. #2
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    Quote Originally Posted by Nivek View Post
    State the possible number of positive real zeros, negative real zeros, and imaginary zeros of each function.

    p(x) = x^5 - 6x^4 - 3x^3 + 7x^2 - 8x + 1

    h(x) = 4x^3 - 6x^2 + 8x - 5

    f(x) = x^3 - 6x^2 + 1

    I would really love to find out how to solve these problems step by step.
    To some extent what you can say about the location of the roots of
    polynomials depends on how much work you do. But I will assume that
    you are expected to use Descartes rule of signs for these.

    I will do the first one.

    Look at the signs of the coefficients:

    + - - + - +

    These change from + to - or - to + 4 times, so by DesCartes rule of
    signs this can have either 4, 2 or 0 positive real roots.

    Now replace x by -x in this to obtain:

    -x^5 - 6x^4 + 3x^3 + 7x^2 + 8x + 1

    so now we have:

    - - + + + +

    so there is 1 change from + to - or - to +, so in this case there must be
    exactly 1 negative real root, (because if there are k changes in the signs
    there can be k, k-2, ... , q, where q is zero if k is even and 1 if k is odd).

    Now complex roots of a polynomial with real coefficients occur in conjugate
    pairs, and the total number of roots is equal to the order of the polynomial
    so here the order is 5 so there are a total of 5 roots altogether.

    So here we have 1 negative root, 4, 2 or 0 positive real roots, with 0, 2 or 4
    complex roots depending on the number positive roots. So in list form we
    have the following possibilities:

    1 negative, 4 positive 0 complex
    1 negative, 2 positive 2 complex
    1 negative, 0 positive 4 complex

    The other problems are similar

    RonL
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  3. #3
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    so there is 1 change from + to - or - to +, so in this case there must be
    exactly 1 negative real root, (because if there are k changes in the signs
    there can be k, k-2, ... , q, where q is zero if k is even and 1 if k is odd).
    Can you re-phrase that?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Nivek View Post
    Can you re-phrase that?
    Try this or this

    RonL
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