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Thread: Factoring Help

  1. #1
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    Factoring Help

    Hi,

    For my FST project, I am supposed to factor n^3 - 8 = 0

    So,

    I turn it into the difference of cubes:

    (n)^3 - (2)^3

    Which ends up being:

    (n - 2) (n^2 + 2n + 4)

    So, (n-2) is one factor. However, I don't know how to find the other zeros. I need to factor (n^2 + 2n + 4), but I can't seem to figure it out.

    Thanks!
    Ethan
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by EthanDavis View Post
    Hi,

    For my FST project, I am supposed to factor n^3 - 8 = 0

    So,

    I turn it into the difference of cubes:

    (n)^3 - (2)^3

    Which ends up being:

    (n - 2) (n^2 + 2n + 4)

    So, (n-2) is one factor. However, I don't know how to find the other zeros. I need to factor (n^2 + 2n + 4), but I can't seem to figure it out.

    Thanks!
    Ethan
    That's good. There are no real factors. You'll have to use the quadratic formula to get the two complex roots.

    -Dan
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  3. #3
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    Re

    Alright, let me try this:

    (n^2 + 2n + 4) =

    -2 +/- root{2^2 - 4(1)(4)
    ______________________
    2

    This equals:

    -2 +/- root{-12
    _____________
    2

    This equals:

    -2 +/- i root{12
    _____________
    2


    This equals:

    -2 +/1 2i root {3
    ______________
    2


    Question: Now is it okay to divide everything by 2?

    -1 +/- i root{3


    So, would those be my factors? (-1+ i root{3)(-1 - i root{3)

    I have a strong feeling I went wrong somewhere along the line...

    (By the way, I apologize for having to write out "+/-" and "root{". How do you draw out the math problem like so many on this forum do?)

    Thanks,
    Ethan
    Last edited by EthanDavis; Apr 17th 2008 at 05:26 PM. Reason: Simplified "root of 12" as "2 roots of 3"
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by EthanDavis View Post
    Alright, let me try this:

    (n^2 + 2n + 4) =

    -2 +/- root{2^2 - 4(1)(4)
    ______________________
    2

    This equals:

    -2 +/- root{-12
    _____________
    2

    This equals:

    -2 +/- i root{12
    _____________
    2


    This equals:

    -2 +/1 2i root {3
    ______________
    2


    Question: Now is it okay to divide everything by 2?

    -1 +/- i root{3


    So, would those be my factors? (-1+ i root{3)(-1 - i root{3)

    I have a strong feeling I went wrong somewhere along the line...

    (By the way, I apologize for having to write out "+/-" and "root{". How do you draw out the math problem like so many on this forum do?)

    Thanks,
    Ethan
    Those are your roots. Your factors, if you choose to write them out, will be
    n^3 - 8 = (n - 2)(n - (-1 + i \sqrt{3} ))(n - (-1 - i \sqrt{3}))

    -Dan
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