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sara1234

how you would analyze the zeros of the polynomial function f(x) = –3x^5 – 8x^4 +25x^3 – 8x^2 +x – 19 using Descartes’ Rule of Signs

  • one year ago
  • one year ago

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  1. sara1234
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    I don't understand what it's asking for is it asking to find positive negitive and complex?

    • one year ago
  2. ash2326
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    @sara1234 It's asking for no. of positive or negative zeros.!!

    • one year ago
  3. sara1234
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    huh? for no of?

    • one year ago
  4. ash2326
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    Yeah, Using Descartes' Rule of Signs we can determine that. I'll explain you

    • one year ago
  5. ash2326
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    We have the polynomial \[f(x) = –3x^5 – 8x^4 +25x^3 – 8x^2 +x – 19\] First step it to count the no. of sign changes in f(x) I'll underline wherever there is a sign change between two consecutive terms i.e from + to - or - to + \[f(x) = –3x^5 \underline{– 8x^4 +25x^3} – 8x^2 +x – 19 \longrightarrow 1st\ change\] \[f(x) = –3x^5 – 8x^4 \underline{+25x^3 – 8x^2} +x – 19 \longrightarrow 2nd\ change\] \[f(x) = –3x^5 – 8x^4 +25x^3 \underline{– 8x^2 +x} – 19 \longrightarrow 3rd\ change\] \[f(x) = –3x^5 – 8x^4 +25x^3 – 8x^2 \underline{+x – 19} \longrightarrow 4th\ change\] so there are 4 sign changes, do you get this part?

    • one year ago
  6. sara1234
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    isnt there 5 sign changes?

    • one year ago
  7. sara1234
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    oh never miind i got it

    • one year ago
  8. ash2326
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    So there could be 4 positive zero, However, some of the roots may be generated by the Quadratic Formula, and these pairs of roots may be complex. so we have to down count by 2. Either it has 4 or 2 or 0 positive roots. Do you get this?

    • one year ago
  9. sara1234
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    yes

    • one year ago
  10. ash2326
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    Now can you write f(-x), it's used to find no. of negative zeros

    • one year ago
  11. sara1234
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    -3(-x)^5-8(-x)^4+25^3-8(-x)^2+(-x)-19

    • one year ago
  12. ash2326
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    Good , we need to simplify it. I'll write that \[f(-x)=-3(-x)^5-8(-x)^4+25(-x)^3-8(-x)^2+(-x)-19\] \[f(-x)=3x^5-8x^4-25x^3-8x^2-x-19\] do you get this?

    • one year ago
  13. sara1234
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    Yes i do

    • one year ago
  14. ash2326
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    Now count the no. of sign changes here

    • one year ago
  15. sara1234
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    only one

    • one year ago
  16. ash2326
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    Yeah so it's having 1 negative zero, we can't down count by 2, it'll become -1. So it has 1 negative zero and 4 or 2 or 0 positive zeros. Do you get this?

    • one year ago
  17. sara1234
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    Yeah thankssss!! Now we find complex?

    • one year ago
  18. ash2326
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    We found that there are either 4 or 2 or 0 positive roots. But these could also be complex, so if we have have 4 positive roots there will be 0 complex if we have 2 positive there will be 2 complex or if we have 0 positive there will be 4 complex

    • one year ago
  19. sara1234
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    why do we have to add the ) positive?

    • one year ago
  20. sara1234
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    0

    • one year ago
  21. ash2326
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    Because 0 could be due to 4 complex roots

    • one year ago
  22. sara1234
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    theres 4 or 2 positive real zeros but why do we have to add the 0

    • one year ago
  23. ash2326
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    There could be a double quadratic, with 4 roots complex. therefore we won't have any positive zero

    • one year ago
  24. sara1234
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    thanks for all your help @ash2326 you really helped me :)

    • one year ago
  25. ash2326
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    you're welcome. did you understand all the things?

    • one year ago
  26. sara1234
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    Yessss :D

    • one year ago
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