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Brunette

According to the rational root theorem, which is not a possible rational root of x^3 + 8x^2 - x - 6 = 0? 1 4 2 6

  • one year ago
  • one year ago

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  1. Mertsj
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    The possible rational roots are the factors of the constant term divided by the factors of the coefficient of the leading term.

    • one year ago
  2. Mertsj
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    The constant term is -6 The leading term is x^3 and its coefficient is 1

    • one year ago
  3. Mertsj
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    So the possible rational roots are the factors of -6 divided by the factors of 1

    • one year ago
  4. Mertsj
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    So what are the possible rational roots?

    • one year ago
  5. Brunette
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    6

    • one year ago
  6. Mertsj
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    What are the factors of -6?

    • one year ago
  7. Brunette
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    no clue???

    • one year ago
  8. Mertsj
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    Let me give you an example: The factors of 12 are 4 and 3 because 4 x 3 = 12

    • one year ago
  9. Mertsj
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    Also 2 and 6 are factors of 12 because 2 x 6 = 12

    • one year ago
  10. Mertsj
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    Also -1 and -12 are factors of 12 because -1 x -12 = 12

    • one year ago
  11. Mertsj
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    Now you tell me some factors of -6

    • one year ago
  12. gojani91
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    what two numbers when multiplied equal 6

    • one year ago
  13. gojani91
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    -6

    • one year ago
  14. Brunette
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    1

    • one year ago
  15. Mertsj
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    ok. Now gojani91 wants to help you. Good bye

    • one year ago
  16. Brunette
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    lol ok then

    • one year ago
  17. gojani91
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    damn you dont have to be rude mertsj

    • one year ago
  18. Brunette
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    lol its fine dont worry about it :)

    • one year ago
  19. Brunette
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    forget it, it's cool ill figure it out but thanks for trying to help i can be quite difficult...

    • one year ago
  20. gojani91
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    what are the factors of -6

    • one year ago
  21. gojani91
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    -1 x 6 1 x -6 2 x -3 3 x -2

    • one year ago
  22. Brunette
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    1?

    • one year ago
  23. Brunette
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    and 2

    • one year ago
  24. Brunette
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    -3 am i right?

    • one year ago
  25. gojani91
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    yes those are all roots

    • one year ago
  26. Brunette
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    k so which do I choose from 1 and 2?

    • one year ago
  27. gojani91
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    this is an easy problem because the leading coefficient is 1 from x^3 so all those numbers can be divided by 1

    • one year ago
  28. gojani91
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    your question ask which is not a possible root number

    • one year ago
  29. Brunette
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    oh ok so 4

    • one year ago
  30. gojani91
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    so what numbers is not a factor of 6

    • one year ago
  31. gojani91
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    yes you got it

    • one year ago
  32. Brunette
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    thank you so much really appreciate it :)

    • one year ago
  33. gojani91
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    no problem

    • one year ago
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