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mskyeg Group Title

Find any rational roots of P(x). P(x)=3x^4+2x^3-9x^2+4

  • 2 years ago
  • 2 years ago

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  1. satellite73 Group Title
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    try 1, -1 first then 2, -2 then 4,-4 if these don't work we have to move on to fractions

    • 2 years ago
  2. mskyeg Group Title
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    there is more than 1 so even if they work there wtill might be a fraction. Oh and with this i have to show my work and do it the textbook way not by trial and error

    • 2 years ago
  3. Freyes Group Title
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    So what exactly is the textbook way for your textbook

    • 2 years ago
  4. satellite73 Group Title
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    there is no other way to do it you have to check. if that is "trial and error' that is what you need to do

    • 2 years ago
  5. Freyes Group Title
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    Well, in practicality my teacher told me to graph the function.. and then you can use synthetic division for the ones that aren't "exact"

    • 2 years ago
  6. Freyes Group Title
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    But unless you have some sort of sixth sense of math you really don't have a set method.. unless your textbook does, in which case i'd want to know.

    • 2 years ago
  7. mskyeg Group Title
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    ya thats the way i am just having trouble relating it to this specifc problem

    • 2 years ago
  8. satellite73 Group Title
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    1 works by your eyeballs because if \[P(x)=3x^4+2x^3-9x^2+4\] then \[P(1)=3+2-9+4=0\]

    • 2 years ago
  9. Freyes Group Title
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    well for this one i would guess -2

    • 2 years ago
  10. Freyes Group Title
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    but its just a guess

    • 2 years ago
  11. satellite73 Group Title
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    add up the coefficients and see that you get 0 then you can factor as \[(x-1)\times (something)\] and then you can either factor the "something" or else use the quadratic formula

    • 2 years ago
  12. satellite73 Group Title
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    you only need to find one zero, then factor and end up with a quadratic, and you can always find the zeros of any quadratic

    • 2 years ago
  13. satellite73 Group Title
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    scratch that, this is a polynomial of degree 4, so you factor and end up with a cubic when you do that, you will see that 1 is a zero of the cubic polynomial as well, so you will factor again as \((x-1)(x-1)(something)\)

    • 2 years ago
  14. Freyes Group Title
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    x+2

    • 2 years ago
  15. Freyes Group Title
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    but there really isn't an exact method to it is all.

    • 2 years ago
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spraguer (Moderator)
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