Results 1 to 3 of 3

Thread: Clueless, a little help needed.

  1. #1
    Oct 2012

    Clueless, a little help needed.


    I've been doing some math recently in school and I'm really confused as to how to work a few things out. I've been asked to 'factorise by extraction and grouping of a common factor from expressions with 2,3 and 4 terms respectively'. And I need help with some of the questions. If someone could be kind enough to include the answers plus how you worked it out.

    Find the factors of:

    1) cd + 6ce

    2) ax + bx

    3) m2n - 2mn2

    4) 3x4y + 9x3y2 - 6x2y3

    5) x2y2 - axy + bxy2

    6) 5a + ax - 2b + by

    7) ax + by + bx + ay

    8) 6x2 + x - 15

    9) 14x2 - 29x + 12

    10)x2 - 16

    Last edited by syther21; Oct 21st 2012 at 04:51 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Dec 2011
    St. Augustine, FL.

    Re: Clueless, a little help needed.

    This site is not a place where you drop off your homework and come back later to find it all completed with work clearly shown.

    We will be glad to help you though.

    Let's look at the first problem. What is the largest factor common to both terms?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Jul 2012

    Re: Clueless, a little help needed.

    If you are referring to x with a power of 2, so x squared, please use the ^ sign which symbolises that the power is higher than the number/algebraic figure. So x squared would be x^2. makes it less confusing, otherwise it looks a bit like 2x.

    And for the questions, remember you are trying to extract the highest factor if I perceived your task correctly. So if you were to do ab + ac, a is common in them both, so you would take out a and leave the rest in brackets, so a(b+c). If you expand this, which means you multiply what is outside of the brackets by what is inside, you will get a*b which is ab (remember the multiplication sign is hidden, so xy = x*y), and a*c which is ac. So in essence try to take out what is the largest number or figure or both, which is common to them, put it outside of the bracket and find what you would need to multiply it with to get what you had before you took out the factor.
    6ab + 4ac =
    2 is the highest number that that fits into 6 and 4 so take out 2. =
    A is common in both. =
    What do you need to multiply 2a by to get 6ab + 4ac? =
    2a(3b + 2c)

    if i understood your question correctly that is
    Last edited by Ashir; Oct 22nd 2012 at 10:21 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Absolutely clueless...
    Posted in the Algebra Forum
    Replies: 13
    Last Post: Jan 20th 2011, 06:12 AM
  2. Optimizations... Clueless
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 9th 2010, 05:04 PM
  3. Clueless on this proof, please help!!!
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: Jun 21st 2010, 11:02 AM
  4. Clueless
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Nov 24th 2009, 08:20 AM
  5. 3x3 eigenvectors. Clueless.
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Apr 6th 2008, 08:45 AM

Search Tags

/mathhelpforum @mathhelpforum