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Math Help - Polynomial function with imaginary root

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    Polynomial function with imaginary root

    The polynomial 3x^3-13x^2+ax+b=0, a and b real numbers, has 2-i as one of its roots. Where does the graph of y=3x^3-13x^2+ax+b cross the x-axis?

    The solution is 1/3. Through what method(s) could i find that answer?
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  2. #2
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    Quote Originally Posted by ceinstein1 View Post
    has 2-i as one of its roots.
    One of the other roots is its complex conjugate 2+i, then use these with the fact that if P(a) = 0 then (x-a) is a factor.

    Also if you want to graph this you with have to use a complex plane with an imaginary and real axis.
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  3. #3
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    If 2-i is a root of f(x) = 3x^3-13x^2+ax+b, then so is 2+i.
    (x-[2-i])(x-[2+i]) = x^2-4x+5, so we can write
    3x^3-13x^2+ax+b = (x^2-4x+5)(cx+d) for some c, d \in\mathbb{R}.
    By expanding the RHS, or otherwise, we see from the coefficient of x^3 that c = 3,
    and from that of x^2 we have d-4c = -13, and as c = 3, we have d = -1.
    Thus it crosses the x-axis at [LaTeX ERROR: Convert failed]
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