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Math Help - Polynomials and x+i

  1. #1
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    Polynomials and x+i

    I could use some help with this:

    Find a polynomial of degree three with real number coefficients and zeros of 3 + i and -7.
    Kind of having trouble with the real number coefficients, and how to find them?
    (x+7)
    (x - (3+1))
    (x + (3-1))
    Is that how you write 3 + i as a zero?

    Any help is greatly appreciated! Thanks!
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  2. #2
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    Quote Originally Posted by Sxon View Post
    I could use some help with this:

    Find a polynomial of degree three with real number coefficients and zeros of 3 + i and -7.
    Kind of having trouble with the real number coefficients, and how to find them?
    (x+7)
    (x - (3+1))
    (x + (3-1))
    Is that how you write 3 + i as a zero?
    Not quite, you write it like this:
    (x + 7)[x - (3 + i)][x - (3 - i)].

    There is a formula regarding multiplying two binomials that contain complex conjugates:
    [x - (a + bi)][x - (a - bi)] = x^2 - 2ax + (a^2 + b^2)

    So
    [x - (3 + i)][x - (3 - i)] = x^2 - 6x + 10.

    Multiply that by x + 7 to get
    (x + 7)(x^2 - 6x + 10) = x^3 + 7x^2 - 6x^2 - 42x + 10x + 70 = x^3 + x^2 - 32x + 70.


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  3. #3
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    Quote Originally Posted by Sxon View Post
    I could use some help with this:

    Find a polynomial of degree three with real number coefficients and zeros of 3 + i and -7.
    Kind of having trouble with the real number coefficients, and how to find them?
    (x+7)
    (x - (3+1))
    (x + (3-1))
    Is that how you write 3 + i as a zero?

    Any help is greatly appreciated! Thanks!

    For x = -7 to be a root, then x + 7 is a factor.

    For 3 + i to be a root, then x - (3 + i) is a factor.

    Since complex solutions always appear as complex conjugates, that means 3 - i is also a solution. So x - (3 - i) is another factor.


    So your polynomial is of the form

    0 = a(x + 7)[x - (3 + i)][x - (3 - i)]

    Expanding gives

    0 = a(x + 7)[x^2 - (3 - i)x - (3 + i)x + (3 + i)(3 - i)]

    0 = a(x + 7)(x^2 - 6x + 4)

    0 = a(x^3 - 6x^2 + 4x + 7x^2 - 42x + 28)

    0 = a(x^3 + x^2 - 38x + 28)


    This equation will hold true for any value of a.

    For simplicity, let a = 1 and then you have your cubic.
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  4. #4
    MHF Contributor red_dog's Avatar
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    If x_1=3+i is a root of the polynomial then x_2=3-i is also a root.

    Then P(x)=(x-3-i)(x-3+i)(x+7)=x^3+x^2-32x+70
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