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HelpMe94 Group Title

Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function: f(x)= -5x^3+X^2-X+5

  • 2 years ago
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  1. mathmate Group Title
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    How many times did the expression changed sign?

    • 2 years ago
  2. HelpMe94 Group Title
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    3

    • 2 years ago
  3. HelpMe94 Group Title
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    i think

    • 2 years ago
  4. mathmate Group Title
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    There is then a MAXIMUM of 3 positive roots, but it could also be 1.

    • 2 years ago
  5. HelpMe94 Group Title
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    why 1?

    • 2 years ago
  6. mathmate Group Title
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    Now flip the sign of the coefficients of the odd powers. How many times does the new expression change sign?

    • 2 years ago
  7. HelpMe94 Group Title
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    Is that 5X^3+X^2+X+5

    • 2 years ago
  8. mathmate Group Title
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    The number of sign changes is the maximum number of positive roots, including complex. If there are two complex roots (as is the case), then there is only one (real) positive root.

    • 2 years ago
  9. HelpMe94 Group Title
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    ok

    • 2 years ago
  10. HelpMe94 Group Title
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    so the negative right converted right?

    • 2 years ago
  11. mathmate Group Title
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    Yes, the number of sign changes of the new expression (i.e. replace x by -x) gives the MAXIMUM number of negative roots.

    • 2 years ago
  12. HelpMe94 Group Title
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    So for this one there is none for the negative and 1 for the positive

    • 2 years ago
  13. HelpMe94 Group Title
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    so how do you know there is one and not 3 for the positive?

    • 2 years ago
  14. mathmate Group Title
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    Using the Descartes rule of signs, we do know that there is no negative root. But we do not know if there is ONE or THREE positive roots.

    • 2 years ago
  15. HelpMe94 Group Title
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    ok so the answer for positive is 1 and 3 and negative is 0

    • 2 years ago
  16. mathmate Group Title
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    You can find that out by factoring the expression. A cubic has at least one real root, and we know that it is positive in this case.

    • 2 years ago
  17. mathmate Group Title
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    correct.

    • 2 years ago
  18. HelpMe94 Group Title
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    one other question:

    • 2 years ago
  19. HelpMe94 Group Title
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    Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function: f(x)=3X^2+2X^2+x+3

    • 2 years ago
  20. HelpMe94 Group Title
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    how does this work then?

    • 2 years ago
  21. HelpMe94 Group Title
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    the positive looks like there is none and same with the negative?

    • 2 years ago
  22. mathmate Group Title
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    Now, let's see. (1) The expression as is has no change in sign, so what do you conclude? (2) flip the sign of the odd powers to give -3X^3+2X^2-x+3, sign changed 3 times, what do you conclude?

    • 2 years ago
  23. HelpMe94 Group Title
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    wait why is 3X^2 flipped?

    • 2 years ago
  24. mathmate Group Title
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    I supposed you have a typo! It's 3x^3, ... or not?

    • 2 years ago
  25. HelpMe94 Group Title
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    let me see...

    • 2 years ago
  26. HelpMe94 Group Title
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    oh srry your right

    • 2 years ago
  27. HelpMe94 Group Title
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    Use the Descartes's Rule of Signs to determine the possible numbers of positive and negative zeros of the function: f(x)=3X^3+2X^2+x+3

    • 2 years ago
  28. HelpMe94 Group Title
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    ok so the positive is none then the negative is -3X^3+3X^2-X+3, so 2 and 1

    • 2 years ago
  29. mathmate Group Title
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    No positive is correct. Can you repeat the conclusion for negative?

    • 2 years ago
  30. HelpMe94 Group Title
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    i don't understand

    • 2 years ago
  31. HelpMe94 Group Title
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    of so negative is 3 and 1

    • 2 years ago
  32. mathmate Group Title
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    The number of possible roots always differ by an even number because complex roots, if any, come in pairs. 1 and 2 will not be correct. 1 and 3 would be.

    • 2 years ago
  33. HelpMe94 Group Title
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    oh ok

    • 2 years ago
  34. mathmate Group Title
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    Cool!

    • 2 years ago
  35. HelpMe94 Group Title
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    cool?

    • 2 years ago
  36. mathmate Group Title
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    Perhaps more exactly, 1 OR 3.

    • 2 years ago
  37. HelpMe94 Group Title
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    ok thanks that makes sense

    • 2 years ago
  38. mathmate Group Title
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    You're welcome! :)

    • 2 years ago
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