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  1. anonymous
    • 5 years ago
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    thats easy , first find where derivative is zero

  2. anonymous
    • 5 years ago
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    so you know where the function turns around ,

  3. anonymous
    • 5 years ago
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    then compare the values of the function at the turning points

  4. anonymous
    • 5 years ago
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    if you go from positive to negative , or negative to positive function value between two turning points then you must have an intersection with the x axis ( ie a zero )

  5. anonymous
    • 5 years ago
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    if the sign of the function value doesnt change then there are no roots in the interval .

  6. anonymous
    • 5 years ago
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    *no real roots

  7. anonymous
    • 5 years ago
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    find the number of the real roots , then to find the number of complex root you subtract the number of real roots from the degree of the function

  8. dumbcow
    • 5 years ago
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    Or you can use Descartes theorem and count the sign changes in f(x) for positive roots and f(-x) for neg roots positive: 2 or 4 negative: 1 imaginary: 0 or 2

  9. anonymous
    • 5 years ago
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    no

  10. anonymous
    • 5 years ago
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    By the fundamental theorem of algebra, this polynomial will have five roots. Descartes rule of signs has that r has either two or zero positive real roots, and one negative real root. The rational root theorem says that 1 is a root of r, and since this is real and positive, r clearly does not have zero real positive roots - it must have two. So, we have so far: Two real positive roots One real negative root Since r has rational coefficients, the remaining roots are complex conjugates by the complex conjugate root theorem. So for r: Two real, positive roots One real, negative root Two complex conjugate roots

  11. anonymous
    • 5 years ago
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    thank all

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