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Math Help - Cubic Polynomial

  1. #1
    Member rtblue's Avatar
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    Cubic Polynomial

    A Cubic Polynomial with integral coefficients has a root (3-4i) and possesses the form:

    y=x^3+ax+b Find AB/25

    Have no clue where to start with this one. Any help is appreciated.
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  2. #2
    A Plied Mathematician
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    If you know one root, you should be able to factor the cubic. One way to do that is by polynomial long division. The result should look like your y.
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  3. #3
    Member rtblue's Avatar
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    How can I factor the polynomial, if the A and B values are not given? On another note, If one of the roots of the polynomial is (3-4i), then another root has to be (3+4i). multiplication of the two gives us

    x^2-6x+25

    Now where do I go from here?
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  4. #4
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    You found that well.

    Now just do this (x-(3-4i))(x-(3+4i) or ((x-3)+4i)((x-3)-4i)=(x-3)^2-(4i)^2

    Regards.
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Well, I tried using the factor theorem.

    Let f(x) = x^3 + ax + b

    Since 3-4i is a factor, f(3-4i) = 0.

    So,  f(3-4i) = (3-4i)^3 + a(3-4i) + b = 0

    Expanding, you get:

    (b - 3a -117) - (44 + 4a)i = 0

    So,

    b - 3a - 117 = 0

    and

    44 + 4a = 0

    I'm not sure if that's right though

    This gives a = -11, and b = 150

    So, \frac{ab}{25} = \frac{(-11)(150)}{25} = -66
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  6. #6
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    Quote Originally Posted by rtblue View Post
    How can I factor the polynomial, if the A and B values are not given? On another note, If one of the roots of the polynomial is (3-4i), then another root has to be (3+4i). multiplication of the two gives us

    x^2-6x+25

    Now where do I go from here?
    Very good. Now, if "c" is the third root, then you must have (x^2- 6x+ 25)(x- c)= x^3+ ax + b.

    (x^2- 6x+ 25)(x- c)= x^3- 6x^2+ 25x- cx^2+ 6cx- 25c= x^3- (6+ c)x^2+ (25+ 6c)x- 25c
    and that must be equal to x^3+ ax+ b

    The fact that there is no " x^2" term on the right tells you that 6+ c= 0 so c= -6. Now, it should be easy to find a and b.
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