Results 1 to 6 of 6

Math Help - Group Factoring

  1. #1
    Member
    Joined
    Jun 2009
    Posts
    86

    Group Factoring

    I have three questions involving group factoring, though I'm not exactly sure how to do them. I've done this so far.

    a)
    ac + bc - ad - bd
    (ac + bc) - (ad - bd)
    c(a + b) - d(a - b)
    (a + b)(c - d)(-1)

    Is the negative supposed to be there?

    b)

    x^2 + 2x + 1 - y^2
    (x^2 + 2x + 1) - y^2
    (x + 1)(x + 1) - y^2
    (x + 1)^2 - y^2

    What do I do from here?

    c)
    x^2 - y^2 - 10y - 25
    x^2 - (y^2 - 10y - 25)
    x^2 - (y + 5)(y - 5)

    Same problem as in b).

    Is what I have even correct so far?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    a) You made a mistake. This is how you should have started:

    ac + bc - ad - bd
    (ac + bc) - (ad + bd)

    b) Good work. Now you can factor the difference of squares:

    (x + 1 - y)(x + 1 + y)
    (x - y + 1)(x + y + 1)

    c) Again, you made a mistake. This is how you should have started:

    x^2 - y^2 - 10y - 25
    x^2 - (y^2 + 10y + 25)

    This one is similar to (b).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2009
    Posts
    86
    a)
    ac + bc - ad - bd
    (ac + bc) - (ad + bd)
    c(a + b) - d(a + b)
    (a + b)(c - d)

    b)

    x^2 + 2x + 1 - y^2
    (x^2 + 2x + 1) - y^2
    (x + 1)(x + 1) - y^2
    (x + 1)^2 - y^2
    (x + 1 - y)(x + 1 + y)
    (x - y + 1)(x + y +1)

    c)
    x^2 - y^2 - 10y - 25
    x^2 - (y^2 + 10y + 25)
    x^2 - (y + 5)(y + 5)
    x^2 - (y + 5)^2
    (x + y + 5)(x - y - 5)

    Not sure about the last line in c). The negative (y + 5)^2 is a little confusing. Got the other two questions. Though, why do the numbers in brackets change to positive?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by shadow6 View Post
    a)
    ac + bc - ad - bd
    (ac + bc) - (ad + bd)
    c(a + b) - d(a + b)
    (a + b)(c - d)

    b)
    x^2 + 2x + 1 - y^2
    (x^2 + 2x + 1) - y^2
    (x + 1)(x + 1) - y^2
    (x + 1)^2 - y^2
    (x + 1 - y)(x + 1 + y)
    (x - y + 1)(x + y +1)

    c)
    x^2 - y^2 - 10y - 25
    x^2 - (y^2 + 10y + 25)
    x^2 - (y + 5)(y + 5)
    x^2 - (y + 5)^2
    (x + y + 5)(x - y - 5)

    Not sure about the last line in c). The negative (y + 5)^2 is a little confusing. Got the other two questions. Though, why do the numbers in brackets change to positive?
    You are correct about (c). To answer your question, take a look at

    x^2 - y^2 - 10y - 25.

    The reason this is equal to

    x^2 - (y^2 + 10y + 25)

    is that the negative sign is "distributed" across all the terms in the sum

    (y^2 + 10y + 25).

    If this is difficult to see, consider that

    x^2 - (y^2 + 10y + 25) is the same as

    x^2 + (-1)(y^2 + 10y + 25)

    and use the distributive law to distribute the -1 across all the terms.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2009
    Posts
    86
    Got it now, thanks.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jun 2009
    Posts
    86
    Got lots more factoring questions today. I managed to complete all of them but 2... The two expressions that need factoring are:

    1) 3x^2-27(2-x)^2

    This one I tried expanding the (2-x)^2, then multiplying the expanded result by 27. I tried this... and I kept getting a much different answer than what was in my textbook. The back of the text book says that the expression's factored form will be -12(2x - 3)(x - 3), though I have absolutely no idea how they got this answer. I've even tried doing this backwards (expanding the answer to get to the question), and that isn't even successful.

    2) 2m^2 + 10m + 10n -2n^2

    This one I've also attempted several tries (in terms of rearranging and grouping, and even factoring out the 2), however I end up getting lost near the end. Got this so far:

    2(m^2 + 5m + 5n - n^2)
    2(m^2 - n^2 + 5m +5n)
    2((m^2 - n^2) + (5m+5n))
    2((m + n)(m + n) + 5(m + n))
    2((m + n) + 5)
    ...?

    Any suggestions?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Factoring by group/grouping
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 10th 2010, 10:21 PM
  2. Replies: 1
    Last Post: November 4th 2009, 10:52 AM
  3. Quick questions on Group Theory - Cosets / Normal Group
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: October 16th 2009, 09:39 AM
  4. Replies: 2
    Last Post: August 22nd 2009, 11:57 AM
  5. Replies: 3
    Last Post: November 6th 2006, 12:02 AM

Search Tags


/mathhelpforum @mathhelpforum