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Math Help - Factoring Polynomial

  1. #1
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    Factoring Polynomial

    I have an old problem that I forgot how I got the answer. The problem is: Find the factored form of a polynomial with real coefficients f(x) that is degree 4, zero's @ 2i and 3 (multiplicity of 2). The function must satisfy f(0) = 72.
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  2. #2
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    Hello,
    Quote Originally Posted by galanm View Post
    I have an old problem that I forgot how I got the answer. The problem is: Find the factored form of a polynomial with real coefficients f(x) that is degree 4, zero's @ 2i and 3 (multiplicity of 2). The function must satisfy f(0) = 72.
    If it's degree 4, then it can be written in this form : f(x)=ax^4+bx^3+cx^2+dx+e

    Then, you have f(2i)=0 ~,~ f(3)=0 ~,~ f(0)=72
    This gives you 3 equations.

    Since 3 is a zero with multiplicity of 2, this means that f'(3)=0 as well

    Now, you only have 4 equations and you need a fifth ?
    Since f is a polynomial with real coefficients, if a complex number is a zero, then its conjugate is also a zero.
    Thus f(-2i)=0

    Looks good to you ?
    Last edited by Moo; June 12th 2009 at 08:41 AM. Reason: 2 not 3
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  3. #3
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    What I have is (x+2i)(x-2i)(x-3)(x-3) the instructor marked it wrong and wrote in a 2 in front of the first term.
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  4. #4
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    I see what I did

    zeros @2i, -2i (conjugate), 3, and 3
    (x+2i)(x-2i)(x-3)(x-3)
    This is what I forgot--SUBSTITUTE 0 for x's gives us 36. Which means we had to multiply the string gy 2 to satisfy f(0)=72.

    Thanks for the sounding board.
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