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sara1234

  • 2 years ago

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

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

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

  3. sara1234
    • 2 years ago
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    huh? for no of?

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

  5. ash2326
    • 2 years ago
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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?

  6. sara1234
    • 2 years ago
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    isnt there 5 sign changes?

  7. sara1234
    • 2 years ago
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    oh never miind i got it

  8. ash2326
    • 2 years ago
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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?

  9. sara1234
    • 2 years ago
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    yes

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

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

  12. ash2326
    • 2 years ago
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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?

  13. sara1234
    • 2 years ago
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    Yes i do

  14. ash2326
    • 2 years ago
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    Now count the no. of sign changes here

  15. sara1234
    • 2 years ago
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    only one

  16. ash2326
    • 2 years ago
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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?

  17. sara1234
    • 2 years ago
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    Yeah thankssss!! Now we find complex?

  18. ash2326
    • 2 years ago
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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

  19. sara1234
    • 2 years ago
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    why do we have to add the ) positive?

  20. sara1234
    • 2 years ago
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    0

  21. ash2326
    • 2 years ago
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    Because 0 could be due to 4 complex roots

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

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

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

  25. ash2326
    • 2 years ago
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    you're welcome. did you understand all the things?

  26. sara1234
    • 2 years ago
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    Yessss :D

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