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Math Help - Finding polynomials

  1. #1
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    Finding polynomials

    Hello, I cant figure this problem out for the life of me. Could someone help me?

    Find the polynomial P(x) of degree 4 with integer coefficients, and zeroes 3-2i and 3 with 3, a zero of multiplicity of 2.



    Thanks in advance...
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  2. #2
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    Quote Originally Posted by Godzilla View Post
    Hello, I cant figure this problem out for the life of me. Could someone help me?

    Find the polynomial P(x) of degree 4 with integer coefficients, and zeroes 3-2i and 3 with 3, a zero of multiplicity of 2.
    Zeros are, therefore, 3-2i, 3+2i, 3, and 3.

    (x - 3 + 2i)(x - 3 - 2i)(x - 3)(x - 3)

    or

    x^4-12x^3+58x^2-132x+117
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  3. #3
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    can someone show me how to factor that? I am confused because of the (3-2i) and (3+2i)? Thanks!!
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  4. #4
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    This polynomial:

    Quote Originally Posted by Haversine View Post
    x^4-12x^3+58x^2-132x+117
    is actually a result of this factorization:

    Quote Originally Posted by Haversine View Post
    (x - 3 + 2i)(x - 3 - 2i)(x - 3)(x - 3)
    That is, if I know the roots are A, B, C, and D, I just set up a series of polynomial factors (x - A)(x - B)(x - C)(x - D).

    There's nothing special about the (x - 3 + 2i) factor, for instance. All it's saying is that I want a term that equals zero when "3 - 2i" is substituted for x.

    The other imaginary term pops in there because the complex conjugate will always be a root.
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  5. #5
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    The whole point is that x^4-12x^3+58x^2-132x+117 is from the product (x-3+2i)(x-3- 2i)(x-3)(x-3). For this problem, at least, it is not a matter of "factoring" but of "multiplying"!

    As far as (x-(3- 2i))(x-(3+2i))= (x-3+2i)(x-3-2i) is concerned, think of it as a "product of sum and difference= difference of squares": ((x-3)- 2i)((x-3)+ 2i)= (x-3)^2- (2i)^2= x^3- 6x+ 9- 4= x^2- 6x+ 5
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