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anonymous

  • 4 years ago

Given a function and one of its zeros, find all the zeros of the function. h(x)=x^3-6x^2+10x-8 given zero: 4

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  1. saifoo.khan
    • 4 years ago
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    I hate when my words fly awaส็็็็็็็็็็็็็็็็็็็็็็็็็็

  2. anonymous
    • 4 years ago
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    we can use synthetic division or just polynomial long division, personally as long as you are dividing by a linear factor, synthetic division is easier to me.

  3. saifoo.khan
    • 4 years ago
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    No help for badman.

  4. anonymous
    • 4 years ago
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    i like synthetic division as well

  5. anonymous
    • 4 years ago
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    Synthethic division, divide the equation by (x - 4) and you get a quadratic equation. Solve that.

  6. anonymous
    • 4 years ago
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    |dw:1327885683777:dw|

  7. anonymous
    • 4 years ago
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    what this means is x^2 - 2x + 2

  8. anonymous
    • 4 years ago
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    okay..and u just solve tht quadratic equation and then u get (x-2) sooo

  9. anonymous
    • 4 years ago
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    It doesnt factor

  10. anonymous
    • 4 years ago
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    oh..srrry..so basically u gotta complete the square?

  11. anonymous
    • 4 years ago
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    or is tht the answer..im still confused

  12. anonymous
    • 4 years ago
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    thats the answer. x^2 - 2x + 2 is one zero, and the other zero is already given

  13. anonymous
    • 4 years ago
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    oh..okay so if it were factorable..tht wooda been the answer?

  14. anonymous
    • 4 years ago
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    kay..ima leave just as promised

  15. anonymous
    • 4 years ago
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    it it was factorable you wouldve factored it and that wouldve been your answer

  16. anonymous
    • 4 years ago
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    kay..thanks :D

  17. anonymous
    • 4 years ago
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    welcome

  18. saifoo.khan
    • 4 years ago
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    badman?

  19. anonymous
    • 4 years ago
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    saying my farewells

  20. saifoo.khan
    • 4 years ago
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    try out the khan-academy vids, it will help u out.

  21. anonymous
    • 4 years ago
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    haha..sure, jus for u, mann..u always find sumthng to make u stand out :P

  22. saifoo.khan
    • 4 years ago
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    ;)

  23. anonymous
    • 4 years ago
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    kay..bye

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