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Math Help - Factoring

  1. #1
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    Factoring

    Factor
    To go about the first one please show me the steps. Thanks
    1. (x+5)^4(x-9)^3+(x+5)^4(x-9)^4
    (x+5)^4(x-9)^3[(1)+(x-9)]?


    2. x^4-15x+144
    Prime?
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  2. #2
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    Re: Factoring

    Quote Originally Posted by theloser View Post
    Factor
    To go about the first one please show me the steps. Thanks
    1. (x+5)^4(x-9)^3+(x+5)^4(x-9)^4
    (x+5)^4(x-9)^3[(1)+(x-9)]?

    This is correct so far, to finish

    (x+5)^4(x-9)^3[(1)+(x-9)]

    (x+5)^4(x-9)^3(x-8)]



    Quote Originally Posted by theloser View Post
    2. x^4-15x+144
    Prime?
    This one looks tricky, are you sure its not x^4-15x^2+144 ??
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  3. #3
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    Re: Factoring

    Quote Originally Posted by pickslides View Post
    This is correct so far, to finish

    (x+5)^4(x-9)^3[(1)+(x-9)]

    (x+5)^4(x-9)^3(x-8)]





    This one looks tricky, are you sure its not x^4-15x^2+144 ??
    O.O I made a mistake...
    It is x^4-15x^2+144
    its not a perfect squared and I don't see any factors of 144 that would total -15.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Factoring

    2. x^4-15x+144
    Prime?
    [/QUOTE]

    x^4>15x-144 for all x in R ===> no factorization over R.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Factoring

    Quote Originally Posted by theloser View Post
    O.O I made a mistake...
    It is x^4-15x^2+144
    its not a perfect squared and I don't see any factors of 144 that would total -15.
    Substitute:

    x^2=t
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  6. #6
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    Re: Factoring

    I think its prime.
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  7. #7
    Super Member Quacky's Avatar
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    Re: Factoring

    See the graph.

    Not that you needed confirmation, but the discriminant (once you've made Zach's substitution) would have told you at a glance.
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