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puppy33

  • one year ago

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

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  1. satellite73
    • one year ago
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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)

  2. satellite73
    • one year ago
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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

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

  4. satellite73
    • one year ago
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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)

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

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

  7. satellite73
    • one year ago
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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

  8. satellite73
    • one year ago
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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"

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

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