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puppy33

help i been try to do this for 2 days i dont understand how do you use Descartes' rule and the Fundamental Theorem of Algebra to predict the number of complex roots of f(x)=x3+x2-x+1 . please explain the process

  • one year ago
  • one year ago

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  1. satellite73
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    fundamental theorem of algebra tells you that there are exactly 3 real zeros of this function, and since it is a cube, either one or all three are complex (a cubic must have on real zero)

    • one year ago
  2. satellite73
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    descartes rule of sign: there are two changes in sign for of the coefficients \[1,1,-1,1\] from 1 to -1 and from -1 to 1, so there are either two positive real zeros or no positive real zeros

    • one year ago
  3. puppy33
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    what would be another good example do show better because im trying to get a good understanding

    • one year ago
  4. satellite73
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    \[f(x)=x^3+x^2-x+1 \] \[f(-x)=-x^3+x^2+x+1 \] and here there is only one change in sign, so there there must be one negative real zero (because you count down by twos, so there must be one)

    • one year ago
  5. satellite73
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    so there could be a) one negative, two positive or b) one negative, two complex

    • one year ago
  6. satellite73
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    here is a really good explanation with worked out examples http://www.purplemath.com/modules/drofsign.htm

    • one year ago
  7. satellite73
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    of course in the modern world we simply check what is right, so this is more of a theoretical (historical) exercise http://www.wolframalpha.com/input/?i=x3%2Bx2-x%2B1 as you can see there is one negative zero and two complex zeros

    • one year ago
  8. satellite73
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    descartes rule of sign does not give you the exact answer. in this example it only tells you "two positive, one negative no complex" or "one negative, two complex"

    • one year ago
  9. puppy33
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    ok i get that part i just dont understand what i have to do with the fundamental theorem of algebra

    • one year ago
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