Math

Write the equation of lowest degree with real coefficients if two of its roots are -1 and 1+i.
The answer is x^3-x^2+2=0. I don't understand why.

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asked by Bethany
  1. complex roots always come in conjugate pairs, so if one is 1+i, you must have another as 1-i

    so we know the equation is
    (x+1)(x - (1+i))(x + 1+i) = 0
    (x+1)(x-1-i)(x+1+i) = 0
    (x+1)(x^2 + x + xi -x -x -i -xi - i - x^2) = 0
    (x+1)(x^2 - 2x + 2) = 0
    x^3 - x^2 + 2x + x^2 - 2x + 2 = 0
    x^3 - x^2 + 2=0

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    posted by Reiny
  2. I did this on paper first, but then messed up when I typed it.

    the first few lines in the equations should have been:
    (x+1)(x - (1+i))(x - (1-i)) = 0
    (x+1)(x-1-i)(x-1+i) = 0
    (x+1)(x^2 - x + xi -x +1 -i -xi + i - i^2) = 0
    the rest is fine

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    posted by Reiny

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