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Math Help - College Alg. Zeros of Polynomial functions

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    College Alg. Zeros of Polynomial functions

    Find a polynomial function of lowest degree with rational coefficients that has (1+i) and 2 as some of it's zeros.
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    Quote Originally Posted by bailey526 View Post
    Find a polynomial function of lowest degree with rational coefficients that has (1+i) and 2 as some of it's zeros.
    Hi bailey526,

    Remember:

    If 2 is a zero, then (x - 2) is a factor.

    Imaginary zeros come in conjugate pairs, so 1 - i is a zero and 1 + i is a zero.

    Therefore, (x - (1 - i)) is a factor and (x - (1 + i)) is a factor.

    Set these factors to zero and expand to find your equation.

    (x - 2)(x - (1 - i))(x - (1 + i)) = 0
    Last edited by masters; April 2nd 2009 at 12:42 PM.
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    Quote Originally Posted by bailey526 View Post
    Find a polynomial function of lowest degree with rational coefficients that has (1+i) and 2 as some of it's zeros.
    \left(x-2\right)\left(x-[1-i]\right)\left(x-[1+i]\right), WHY?
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    So, I did
    0 = (x-2) (1+i) (1-i)
    and then got it to 0 = (x-2)
    x=2...
    Is that correct and what do I do with that number? And I don't quite understand how it says "polynomial function of the lowest degree"
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    Quote Originally Posted by bailey526 View Post
    So, I did
    0 = (x-2) (1+i) (1-i)
    and then got it to 0 = (x-2)
    x=2...
    Is that correct and what do I do with that number? And I don't quite understand how it says "polynomial function of the lowest degree"
    No it is certainly not correct!
    You do know that \left(1-i\right)\left(1+i\right)=2,

    Then \left(x-2\right)\left(x-[1-i]\right)\left(x-[1+i]\right)= x^3 -4x^2 +6x -4.
    Last edited by Plato; April 3rd 2009 at 04:06 AM.
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    Quote Originally Posted by Plato View Post
    No it is certainly not correct!
    You do know that \left(1-i\right)\left(1+i\right)=2,

    Then \left(x-2\right)\left(x-[1-i]\right)\left(x-[1+i]\right)= {\color{red}x^3} -4x^2 +6x -4.
    Typo.
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    Quote Originally Posted by bailey526 View Post
    So, I did 0 = (x-2) (1+i) (1-i)....
    It looks like you might not be familiar with the concepts involved here...?

    Think about when you solved quadratic equations by factoring. You'd get a factored equation that looked something like (x - a)(x - b) = 0, and you'd solve by setting each factor equal to zero: x - a = 0 or x - b = 0, so x = a or x = b.

    Now, you need to work backwards. Given that x = a and x = b are zeroes, they must have solved x - a = 0 and x - b = 0, so x - a and x - b must have been factors.

    By multiplying these factors, you can find the quadratic from its zeroes.

    Note, however, that 3(x - a)(x - b) has the same zeroes (since the "3" will be divided off). Note also that (x - a)^3 (x - b)^4 has the same zeroes, but is of higher degree.

    To find "the polynomial of least degree" with a given set of zeroes, you don't over-duplicate like I just did above.

    For worked examples of how this process works with polynomials (not just quadratics) and complex roots, try here.

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