Results 1 to 7 of 7

Math Help - Factoring Expressions raised to powers

  1. #1
    Member
    Joined
    May 2008
    Posts
    143

    Factoring Expressions raised to powers

    Question 1 solved

    Question 2 I have a question for the expression

    when the author factors out they leave the term the same resulting in the answer
    ...doesn't this term need to be "broken up" like all of the other terms once its factored?

    Many Thanks
    Last edited by allyourbass2212; July 1st 2009 at 04:52 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

    Didn't you understand the previous clarification ?
    http://www.mathhelpforum.com/math-he...ed-powers.html
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    143
    Quote Originally Posted by running-gag View Post
    Hi

    Didn't you understand the previous clarification ?
    http://www.mathhelpforum.com/math-he...ed-powers.html
    Not completely regarding the question in this topic, thus the reason for making a new thread and not bumping the previous.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,110
    Thanks
    982
    Quote Originally Posted by allyourbass2212 View Post


    In the expression

    2(x^2-6)^9+(x^2-6)^4+4(x^2-6)^3+(x^2-6)^2
    =[2(x^2-6)^7+(x^2-6)^2+4x^2-24{\color{red} +1}](x^2-6)^2

    I.
    a
    .Where is the +1 coming from?
    b.I am assuming they are using the coefficient from + (x^2-6)^2
    c.If this is correct then why does the +1 get grouped within  x^2-24 resulting in -23, rather staying on the outside of the brackets? e.g.
    =[2(x^2-6)^7+(x^2-6)^2+4x^2-24]{\color{red} +1}(x^2-6)^2

    II.
    And if thats the case why are they not using a +1 in the second term as well?
    =[2(x^2-6)^7{\color{red} +1}(x^2-6)^2+4x^2-24](x^2-6)^2

    Many Thanks for clarification!
    2(x^2-6)^9+(x^2-6)^4+4(x^2-6)^3+(x^2-6)^2

    let u = x^2-6

    2u^9+u^4+4u^3+u^2

    factor out u^2 ...

    u^2(2u^7 + u^2 + 4u + 1) ... see the "1" ?

    back substitute ...

    (x^2-6)^2[2(x^2-6)^7 + (x^2-6)^2 + 4(x^2-6) + 1]<br />

    distribute the 4(x^2-6) term ...

    (x^2-6)^2[2(x^2-6)^7 + (x^2-6)^2 + 4x^2 - 24 + 1]

    (x^2-6)^2[2(x^2-6)^7 + (x^2-6)^2 + 4x^2 - 23]
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2008
    Posts
    143
    I have a question for the expression

    when the author factors out they leave the term the same resulting in the answer
    ...doesn't this term need to be "broken up" like all of the other terms once its factored?

    Many Thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2008
    Posts
    143
    Bump, would someone please help?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,110
    Thanks
    982
    Quote Originally Posted by allyourbass2212 View Post
    I have a question for the expression

    when the author factors out they leave the term the same resulting in the answer
    ...doesn't this term need to be "broken up" like all of the other terms once its factored?

    Many Thanks
    (15xy-1)(2x-1)^3 - 8(2x-1)^2

    let u = (2x-1)

    (15xy-1)u^3 - 8u^2

    factor out u^2 ...

    u^2[(15xy-1)u - 8]

    back substitute (2x-1) for u ...

    (2x-1)^2[(15xy-1)(2x-1) - 8]

    note that it is permissible to stop here, or one can expand the binomial product in the brackets and combine the resulting constant terms ...

    (2x-1)^2[(30x^2y - 15xy - 2x + 1) - 8]<br />

    (2x-1)^2(30x^2y - 15xy - 2x - 7)

    really no big advantage in doing this, however.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Factoring expressions
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 24th 2010, 07:48 PM
  2. factoring expressions
    Posted in the Algebra Forum
    Replies: 7
    Last Post: January 13th 2010, 12:12 PM
  3. Factoring expressions
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 1st 2009, 11:10 PM
  4. Factoring expressions raised to powers
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 19th 2009, 12:40 PM
  5. Replies: 4
    Last Post: March 18th 2009, 06:27 AM

Search Tags


/mathhelpforum @mathhelpforum