Results 1 to 5 of 5

Math Help - Factoring completely

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    Lagendia
    Posts
    6

    Factoring completely

    how to factor this completely

    6a^2+12ab+6b^2+7ac+7bc-20c^2?
    thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,021
    Thanks
    280

    Re: Factoring completely

    (2a+2b+5c)(3a+3b-4c)

    It's a matter of trial and error, but there are few clues you can use. First notice that the coefficients for all the a and b terms are the same - that tells you that the coefficients for a and b in the factored form are most likely the same as well. The fact that all coeefficients are positive except for the c^2 term tells you that all coefficients of the factored terms are positive except for one of the c's. The factors of 6 are either 1 and 6 or 2 and 3, so try both and see where it leads.
    Last edited by ebaines; August 9th 2012 at 12:08 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,313
    Thanks
    1291

    Re: Factoring completely

    What do you mean by "factor completely"? What I see is that 6^2+ 12ab+ 6b^2= 6(a^2+ 2ab+ b^2)= 6(a+ b)^2 and that 7ac+ 7bc= 7c(a+ b). That gives 6a^2+ 12ab+ 6b^2= 6(a+ b)^2+ 7c(a+ b)- 20c^2. I would not consider that "factored completely" but that is the best I can do.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2012
    From
    Lagendia
    Posts
    6

    Re: Factoring completely

    Quote Originally Posted by ebaines View Post
    (2a+2b+5c)(3a+3b-4c)

    It's a matter of trial and error, but there are few clues you can use. First notice that the coefficients for all the a and b terms are the same - that tells you that the coefficients for a and b in the factored form are most likely the same as well. The fact that all coeefficients are positive except for the c^2 term tells you that all coefficients of the factored terms are positive except for one of the c's. The factors of 6 are either 1 and 6 or 2 and 3, so try both and see where it leads.
    Nice! Thanks. But can you show the solution? I didnt get the 7ac+7bc-20c^2 part there.


    Quote Originally Posted by HallsofIvy View Post
    What do you mean by "factor completely"? What I see is that 6^2+ 12ab+ 6b^2= 6(a^2+ 2ab+ b^2)= 6(a+ b)^2 and that 7ac+ 7bc= 7c(a+ b). That gives 6a^2+ 12ab+ 6b^2= 6(a+ b)^2+ 7c(a+ b)- 20c^2. I would not consider that "factored completely" but that is the best I can do.
    Uhm. I think Factoring completely is, all answers must be form in a parentheses, no other operations. Thanks for help and solution but I need to know how to factor that 7ac+7bc-20c^2. (
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,021
    Thanks
    280

    Re: Factoring completely

    Quote Originally Posted by gashbell1 View Post
    Nice! Thanks. But can you show the solution? I didnt get the 7ac+7bc-20c^2 part there.
     (2a+2c+5c)(3a+3b-4c) = 2a(3a+3b-4c)+2b(3a+3b-4c)+5c(3a+3b-4c) =6a^2+6ab-8ac+6ab+6b^2-8bc+15ac+15bc-20c^2 = 6a^2+12ab+6b^2+7ac+7bc-20c^2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Factoring completely.
    Posted in the Algebra Forum
    Replies: 5
    Last Post: July 3rd 2011, 11:06 PM
  2. Factoring Completely
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 7th 2009, 04:35 PM
  3. Factoring Completely
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 31st 2009, 05:35 AM
  4. Factoring Completely...
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 18th 2008, 09:53 PM
  5. factoring completely
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 11th 2007, 08:00 PM

Search Tags


/mathhelpforum @mathhelpforum