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anonymous

  • one year ago

Help please :) List the possible zeros for the polynomial. Also, use Descartes' rule of signs to describe possible combinations of roots for the polynomials. h(x)=2x^3+5x^2-31x-15

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  1. anonymous
    • one year ago
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    how many sign changes are there for h(x) how many sign changes are there for h(-x)

  2. anonymous
    • one year ago
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    one sign change for h(x) and im not sure about h(-x)

  3. anonymous
    • one year ago
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    h(-x)=2(-x)^3+5(-x)^2-31(-x)-15

  4. anonymous
    • one year ago
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    is it also one?

  5. anonymous
    • one year ago
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    h(-x)=2(-x)^3+5(-x)^2-31(-x)-15 = -2x^3 + 5x^2 + 31x - 15

  6. anonymous
    • one year ago
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    i see two sign changes

  7. anonymous
    • one year ago
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    oh, oops! now what?

  8. triciaal
    • one year ago
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    the first part of the question find the possible zeros

  9. anonymous
    • one year ago
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    The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number. Multiple roots of the same value are counted separately.

  10. anonymous
    • one year ago
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    By descartes law of signs there must be 1 positive root and two or zero negative roots.

  11. anonymous
    • one year ago
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    how do i find the combinations after knowing that?

  12. anonymous
    • one year ago
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    the possible rational roots are of the form p/q where p divides the last coefficient, and q divides the first coefficient

  13. anonymous
    • one year ago
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    so would it be -15/2

  14. anonymous
    • one year ago
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    $$ \Large \frac{p}{q} = \frac{\pm 1, \pm 3, \pm 5, \pm 15}{\pm 1,\pm 2} $$

  15. nincompoop
    • one year ago
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    descartes was not a complete fool just because he did not discover calculus.

  16. triciaal
    • one year ago
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  17. anonymous
    • one year ago
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    thanks @jayzdd :) i get it haha

  18. anonymous
    • one year ago
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    It turns out all the zeroes are irrational.

  19. anonymous
    • one year ago
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    but the number of real roots did obey descartes law of signs

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