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bearsams

  • 2 years ago

find all real zeros f(x)=2x^4+3x^3-6x^2-6x+4

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  1. bearsams
    • 2 years ago
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    anyone there

  2. zzr0ck3r
    • 2 years ago
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    algebraically?

  3. bearsams
    • 2 years ago
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    no

  4. bearsams
    • 2 years ago
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    using synthetic division

  5. zzr0ck3r
    • 2 years ago
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    ahh word. Ill watch to learn:)

  6. bearsams
    • 2 years ago
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    you know how/

  7. zzr0ck3r
    • 2 years ago
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    well we know, all non real roots come in tandem. that's all I can help:)

  8. bearsams
    • 2 years ago
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    iopus?

  9. lopus
    • 2 years ago
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    |dw:1348011701036:dw| this is ruffinni

  10. bearsams
    • 2 years ago
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    thank you so much

  11. bearsams
    • 2 years ago
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    then i have to put it in a quadratic equation

  12. lopus
    • 2 years ago
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    |dw:1348012143081:dw|

  13. bearsams
    • 2 years ago
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    wow ok i understand thanks a bunch

  14. lopus
    • 2 years ago
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    you understand ruffinni? x=-2 x=1/2 x=... x=..

  15. bearsams
    • 2 years ago
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    ruffinni? never heard of it.. but i know how u got it now

  16. lopus
    • 2 years ago
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    ruffini is synthetic division you can finish this. lets go!

  17. bearsams
    • 2 years ago
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    ok hold on

  18. bearsams
    • 2 years ago
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    how is there 2 more x's

  19. lopus
    • 2 years ago
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    |dw:1348012608903:dw|

  20. bearsams
    • 2 years ago
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    so + or - square root of 2 is one x too and that leaves one more?

  21. lopus
    • 2 years ago
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    |dw:1348012735984:dw|

  22. bearsams
    • 2 years ago
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    so the x's are x= +square root of 2 x= -square root of 2 x= 1/2 x=-2

  23. lopus
    • 2 years ago
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    yes

  24. bearsams
    • 2 years ago
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    thank you so much

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