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Math Help - Polynomials

  1. #1
    VMM
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    Polynomials

    What is the degree of the simplest polynomial with the zeroes:√2, 2√2 + i, and 2 + 2i ??
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  2. #2
    Junior Member NowIsForever's Avatar
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    My guess would be 6 since the simplest polynomial that I can come up with that has all of these as roots is (x - 2)(x - 4√2 x + 9)(x - 4x + 8).
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  3. #3
    Super Member Bacterius's Avatar
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    This means that x - \sqrt{2}, x - (2 \sqrt{2} + \pi), x - (2 + 2 \pi) are all factors of this polynomial. Therefore the simplest polynomial that has all these roots is :

    (x - \sqrt{2})(x - (2 \sqrt{2} + \pi))(x - (2 + 2 \pi)) = 0

    Or, equivalently :

    (x - \sqrt{2})(x - 2 \sqrt{2} - \pi)(x - 2 - 2 \pi) = 0

    And the degree of this polynomial is, trivially, 3 (can be checked by expanding).
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  4. #4
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    Quote Originally Posted by VMM View Post
    What is the degree of the simplest polynomial with the zeroes:√2, 2√2 + i, and 2 + 2i ??
    Roots or zeros of polynomials of degree greater than 2 - Topics in precalculus
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