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Math Help - rational zeros

  1. #1
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    rational zeros

    f(x)=x^3-5x^2-14x
    f(x)=x^3+4x^2+9x+36
    f(x)=x^4+x^3-2x^2+4x-24


    i really need the helpp!
    thanks in advance!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by missunique View Post
    f(x)=x^3-5x^2-14x
    simply factor out an x and you obtain a quadratic to factorize.

    f(x)=x^3+4x^2+9x+36
    factor by grouping. there are common terms among the first two terms and the last two terms of the original cubic

    f(x)=x^4+x^3-2x^2+4x-24
    by the rational roots theorem (look it up), if there are rational roots to this equation, they will be from the factors of 24 (namely, \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \mbox{ and } \pm 24). start with the ones and plug them in and see if you get zero. if you obtain 0 for one, then it means that is a root. you can repeat this process with the others. or you may prefer to use polynomial long division or synthetic division to break down the polynomial in the following way.

    let a be one of the roots you found using the above method. so a is a root. by the factor theorem, (x - a) is a factor of the equation. so use long (or synthetic) division to compute (x^4 + x^3 - 2x^2 + 4x - 24) \div (x - a)

    you will obtain a cubic as the result. for which you can repeat the process (or perhaps factoring by grouping can work). this way you keep breaking it down until you get to a quadratic, where factoring (i hope) is easy for you.

    try them, good luck. if you don't understand something, say so
    Last edited by Jhevon; March 23rd 2008 at 12:15 PM. Reason: Thanks CaptainBlack
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  3. #3
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    =]

    ty
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by katyy! View Post
    ty
    hehe, who is this?

    you followed everything ok? i suppose what i wrote for the last question might be a bit confusing to someone with little experience in these kinds of problems, but i just wanted to get the brain juices flowing...
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  5. #5
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    =]

    yea, the last was a bit confusing, but me and my friend think we got it!
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by katyy! View Post
    yea, the last was a bit confusing, but me and my friend think we got it!
    if it's not too much trouble, would you like to post the solution to the last problem? of course, you don't have to show the long (or synthetic) division part. just say where you did it.
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  7. #7
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    this?

    12x^3+31x^2-17x-6/x+3=12x^2-5x-2
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by katyy! View Post
    12x^3+31x^2-17x-6/x+3=12x^2-5x-2
    no. i want you to show how to factorize x^4 + x^3 - 2x^2 + 4x - 24
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    by the rational roots theorem (look it up), if there are rational roots to this equation, they will be from the factors of 24 (namely, \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12).
    What happened to \pm24?

    RonL
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    What happened to \pm24?

    RonL
    yes, i forgot that

    i corrected it
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