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Math Help - n=3; -5 and 4+3i are zeros; f(2) =91

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    n=3; -5 and 4+3i are zeros; f(2) =91

    I'm really hung up on the following part of the problem:

    (x+4+3i)(x-4+3i)

    What I came up with:

    x^2 +4x +3xi -4x +3xi +12i -12i +9i^2 -16: Combine like terms to get

     <br />
x^2 +6xi -25<br />

    I've also tried using -6 in place of 6xi but nope.

    I then multiply by (x+5) to get a host of wrong answers.

    I think my problem here is that I just don't know the rules for what to do with imaginary numbers in this situation. I've internet searched but not found anything satisfactory. There's no comparable problem in the book and other online searches haven't yielded what I'm looking for. Any help is much appreciated. Thank you.
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    Re: n=3; -5 and 4+3i are zeros; f(2) =91

    Quote Originally Posted by Ingersoll View Post
    I'm really hung up on the following part of the problem:

    (x+4+3i)(x-4+3i)

    What I came up with:

    x^2 +4x +3xi -4x +3xi +12i -12i +9i^2 -16: Combine like terms to get

    x^2 +6xi -25

    I've also tried using -6 in place of 6xi but nope.

    I then multiply by (x+5) to get a host of wrong answers.

    I think my problem here is that I just don't know the rules for what to do with imaginary numbers in this situation. I've internet searched but not found anything satisfactory. There's no comparable problem in the book and other online searches haven't yielded what I'm looking for. Any help is much appreciated. Thank you.
    Please, post the exact problem.
    What is going on? What is to be done?
    There is object stated.
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  3. #3
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    Re: n=3; -5 and 4+3i are zeros; f(2) =91

    Quote Originally Posted by Ingersoll View Post
    I'm really hung up on the following part of the problem:

    (x+4+3i)(x-4+3i)

    What I came up with:

    x^2 +4x +3xi -4x +3xi +12i -12i +9i^2 -16: Combine like terms to get

     <br />
x^2 +6xi -25<br />

    I've also tried using -6 in place of 6xi but nope.

    I then multiply by (x+5) to get a host of wrong answers.

    I think my problem here is that I just don't know the rules for what to do with imaginary numbers in this situation. I've internet searched but not found anything satisfactory. There's no comparable problem in the book and other online searches haven't yielded what I'm looking for. Any help is much appreciated. Thank you.
    if 4+3i is a zero, then its conjugate, 4-3i, is also.

    f(x) = k(x+5)[x-(4+3i)][x-(4-3i)]

    f(x) = k(x+5)(x^2-8x+25)

    f(2) = 91 = k(2+5)(4-16+25) = k(7)(13)

    k = 1

    f(x) = (x+5)(x^2-8x+25)
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