Results 1 to 7 of 7

Thread: Factoring

  1. #1
    Junior Member
    Joined
    Jul 2011
    Posts
    61

    Factoring

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


    2.$\displaystyle x^4-15x+144$
    Prime?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33

    Re: Factoring

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

    This is correct so far, to finish

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

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



    Quote Originally Posted by theloser View Post
    2.$\displaystyle x^4-15x+144$
    Prime?
    This one looks tricky, are you sure its not $\displaystyle x^4-15x^2+144$ ??
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2011
    Posts
    61

    Re: Factoring

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

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

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





    This one looks tricky, are you sure its not $\displaystyle x^4-15x^2+144$ ??
    O.O I made a mistake...
    It is $\displaystyle x^4-15x^2+144$
    its not a perfect squared and I don't see any factors of 144 that would total -15.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: Factoring

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

    $\displaystyle x^4>15x-144$ for all x in R ===> no factorization over R.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: Factoring

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

    x^2=t
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jul 2011
    Posts
    61

    Re: Factoring

    I think its prime.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Quacky's Avatar
    Joined
    Nov 2009
    From
    Windsor, South-East England
    Posts
    901

    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need help with factoring
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Feb 7th 2010, 10:47 AM
  2. factoring help
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Feb 4th 2010, 07:44 PM
  3. Is this factoring or something?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Feb 1st 2010, 06:54 PM
  4. Replies: 2
    Last Post: Aug 22nd 2009, 10:57 AM
  5. Replies: 3
    Last Post: Nov 5th 2006, 11:02 PM

Search Tags


/mathhelpforum @mathhelpforum