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Math Help - Complex Zero's of a Function

  1. #1
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    Complex Zero's of a Function

    Hi everyone. First time here and I was wondering if someone could assist me with this math problem that I'm having trouble with.

    The problem reads:

    find all complex zeros of the polynomial function, give exact values, list multiple zeros as necessary

    f(x)= x^4+10x^3-31x^2+80x-312


    I know I'm supposed to do something with rational zero principle or maybe synthetic division but I wasn't sure.

    Any help would be appreciated thanks
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  2. #2
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    Quote Originally Posted by Anandymous View Post
    Hi everyone. First time here and I was wondering if someone could assist me with this math problem that I'm having trouble with.

    The problem reads:

    find all complex zeros of the polynomial function, give exact values, list multiple zeros as necessary

    f(x)= x^4+10x^3-31x^2+80x-312

    I know I'm supposed to do something with rational zero principle or maybe synthetic division but I wasn't sure.

    Any help would be appreciated thanks
    Something worth checking since 312 = 39 \cdot 8:

    x^4 + 10x^3 - 31x^2 + 80x - 312 = (x^4 + 10x^3 - 39x^2) + (8x^2 + 80x - 312) = x^2 (x^2 + 10x - 39) + 8(x^2 + 10x - 39) = \, ....

    Aha!
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    Quote Originally Posted by mr fantastic View Post
    Something worth checking since 312 = 39 \cdot 8:

    x^4 + 10x^3 - 31x^2 + 80x - 312 = (x^4 + 10x^3 - 39x^2) + (8x^2 + 80x - 312) = x^2 (x^2 + 10x - 39) + 8(x^2 + 10x - 39) = \, ....

    Aha!
    Thanks for your help Mr. Fantastic (haha ) but perhaps you could shed some light on your response. I think I somewhat understand where you are going with this, but I am not familiar with this method of solution. Should I be looking at the factors in front of the parenthesis (x^2 and 8) in order to form a solution?
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    Quote Originally Posted by Anandymous View Post
    Thanks for your help Mr. Fantastic (haha ) but perhaps you could shed some light on your response. I think I somewhat understand where you are going with this, but I am not familiar with this method of solution. Should I be looking at the factors in front of the parenthesis (x^2 and 8) in order to form a solution?
    Take out the common factor and you get (x^2 + 8) (x^2 + 10x - 39).

    Now factorise each quadratic factor over C.

    (Once you've done this you'll see that your alternative was to spot x = 3 as a zero and go from there ....)
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    Doh, I'm dumb. So basically, X^2 + 8 solves into X^2 = -8 and then I simply have to solve for X, which is +- sqrt(-8)
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    Quote Originally Posted by Anandymous View Post
    Doh, I'm dumb. So basically, X^2 + 8 solves into X^2 = -8 and then I simply have to solve for X, which is +- sqrt(-8)
    Yes. And feel free to use the symbol  i in an appropriate way (or j if you're an electrical engineer in the making).
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