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jpjones

  • 2 years ago

find all roots of the polynomial p(z) = 2z^3-2z^2-z-6

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  1. vikrantg4
    • 2 years ago
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    cubic equation omg

  2. jpjones
    • 2 years ago
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    z = x+iy this is complex number question

  3. baddinlol
    • 2 years ago
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    Well first of all you can find out that P(2)=0

  4. baddinlol
    • 2 years ago
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    Therefore z-2 must be a factor of 2z^3 - 2z^2 - z - 6

  5. Yahoo!
    • 2 years ago
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    There will be one + ve real root.)

  6. jpjones
    • 2 years ago
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    why is it that you have fed 2 into the equation, is this a random number

  7. baddinlol
    • 2 years ago
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    When you try to factorize polynomials you can try plugging in numbers to see which one of them give you a zero, it is a trial and error way to factorize easily

  8. baddinlol
    • 2 years ago
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    So if P(2) gives you zero, z-2 must be a factor of p(z)

  9. baddinlol
    • 2 years ago
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    Then when you divide 2z^3 + 2z^2-z-6 by z-2

  10. baddinlol
    • 2 years ago
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    You get 2z^2 + 2z +3

  11. Yahoo!
    • 2 years ago
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    Sum of the Roots , product of the roots

  12. baddinlol
    • 2 years ago
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    So we have the following expression (2z^2 +2z +3)(z-2)=0

  13. baddinlol
    • 2 years ago
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    Now we have found one solutions z = 2

  14. baddinlol
    • 2 years ago
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    Other two are given by 2z^2+2z+3=0

  15. Yahoo!
    • 2 years ago
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    and the other sol^n will be Complex numbers...

  16. Yahoo!
    • 2 years ago
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    Gud work...@baddinlol

  17. baddinlol
    • 2 years ago
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    I am going to complete the square

  18. jpjones
    • 2 years ago
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    polynomials are fun aren't they

  19. Yahoo!
    • 2 years ago
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    nop...@baddinlol use quadratic formula

  20. baddinlol
    • 2 years ago
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    2(z^2 + z + 3/2) = 0 z^2 + z + 3/2 = 0 (z + 1/2)^2 + 5/4 = 0

  21. baddinlol
    • 2 years ago
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    (z + 1/2)^2 = -5/4 z + 1/2 = sqrt(5)i /2, -sqrt(5)i /2 z = (sqrt(5)i - 1)/2, (-1-sqrt(5)i)/2

  22. jpjones
    • 2 years ago
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    that will be =+or-sqrt b^2-4(a)(c)/(2a)

  23. jpjones
    • 2 years ago
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    well done baddinol thanks

  24. baddinlol
    • 2 years ago
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    np!

  25. baddinlol
    • 2 years ago
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    Did you know about putting in numbers to find p(Z) = 0?

  26. baddinlol
    • 2 years ago
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    By the way the numbers you plug in have to be a factor of the constant in the equation, which in your equation is -6::::: 2z^3-2z^2-z-6

  27. jpjones
    • 2 years ago
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    i did but i thought there must be a faster way

  28. baddinlol
    • 2 years ago
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    SO i tried 1, -1, 2 and when i got upto 2 i stopped because i found P(2)=0

  29. baddinlol
    • 2 years ago
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    Are you allowed to use a cas calculator?

  30. jpjones
    • 2 years ago
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    no

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