Results 1 to 9 of 9

Math Help - Factoring Questions

  1. #1
    Member
    Joined
    May 2008
    Posts
    143

    Factoring Questions

    1. When factoring by grouping I am having problems understanding a particular step.
    Expression:  2x^4-6x-x^3y+3y

    =(2x^4-6x)-(x^3y+3y)
    =2x(x^3-3)+y(x3-3) Why are we able to change the signs towards the end of the expression? In the first step its a "-" sign and at the next step its a "+" sign.
    =(2x-y)(x^3-3) and in the final solution the sign changes again...

    2. Reducing a fraction with -1
    "Reducing a fraction or adding two fractions sometimes only requires that -1 be factored from one or more denominators"
    Expression:  \frac{3}{y-x}+\frac{x}{x-y}=\frac{-3+x}{x-y}
    -The book lists the steps to solve this but they are rather confusing and they do not clearly explain, I would prefer a member on here run me through the above expression and provide the how/why when solving this problem.
    Last edited by allyourbass2212; June 25th 2009 at 07:51 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    1) If you divide a positive (+3y) by a negative (-y), what should be the sign on the answer?

    2) You have y - x, which equals -x + y, which equals -1(x - y), which then allows you to have a common denominator.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    143
    Quote Originally Posted by stapel View Post
    1) If you divide a positive (+3y) by a negative (-y), what should be the sign on the answer?

    2) You have y - x, which equals -x + y, which equals -1(x - y), which then allows you to have a common denominator.

    im afraid I do not understand your first explanation, the second is more clear to me at the moment.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    May 2009
    Posts
    527
    Quote Originally Posted by allyourbass2212 View Post
    1. When factoring by grouping I am having problems understanding a particular step.
    Expression:  2x^4-6x-x^3y+3y
    =(2x^4-6x)-(x^3y+3y)
    This is wrong. It should be =(2x^4-6x)-(x^3y\;{\color{red}-}\;3y) The reason is that we essentially factored out a -1 in the last two terms.

    =2x(x^3-3)+y(x3-3)
    This should be =2x(x^3-3)\;{\color{red}-}\;y(x^3 - 3)

    =(2x-y)(x^3-3)
    This is correct.


    01
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2008
    Posts
    143
    Quote Originally Posted by yeongil View Post
    This is wrong. It should be =(2x^4-6x)-(x^3y\;{\color{red}-}\;3y) The reason is that we essentially factored out a -1 in the last two terms.
    This should be =2x(x^3-3)\;{\color{red}-}\;y(x^3 - 3)

    Would you mine explaining this in further detail? Because I still dont understand where/why/ and how the -1 is factored out for the last 2 terms .
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    May 2009
    Posts
    527
    Quote Originally Posted by allyourbass2212 View Post
    This is wrong. It should be =(2x^4-6x)-(x^3y\;{\color{red}-}\;3y) The reason is that we essentially factored out a -1 in the last two terms.
    Would you mine explaining this in further detail? Because I still dont understand where/why/ and how the -1 is factored out for the last 2 terms .
    It's because of the parentheses.

    If you take an expression like
    -a - b
    you can factor out a -1. In doing so, you switch the signs inside:
    -1(a + b)
    but you don't need to write the 1 really:
    -(a + b)

    Now if you have an expression like
    -ax + bx
    Factor out a -1x (or -x) and switch the signs inside:
    -1x(a - b) = -x(a - b)

    The original expression was
     2x^4-6x-x^3y+3y

    We're factoring by grouping. Factoring out the common factor from the first two terms is easy enough:
     2x(x^3-3)-x^3y+3y

    We factor out a -1y (or a -y) from the last two terms:
     2x(x^3-3)\;{\color{red}-1y}\;(x^3-3)
     2x(x^3-3)\;{\color{red}-y}\;(x^3-3)


    01
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    May 2008
    Posts
    143
    Many Thanks Kind Sir!

    Yeongil, may I also ask that you clarify my expression 2 question?

    2. Reducing a fraction with -1
    Expression:  \frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-(-y+x)}+\frac{x}{x-y}= \frac{3}{-(x-y)}+\frac{x}{x-y}, now the last step here is where I become completely lost, where the denominator goes from \frac{3}{-(-y+x)} to \frac{3}{-x(x-y)} in the next step in solving the expression

    then the last part I am also lost
     =\frac{3}{-(x-y)}+\frac{x}{x-y}=\frac{-3}{x-y}+\frac{x}{x-y} in the last part of solving this expression how does the once negative denominator of =\frac{3}{-(x-y)} go to \frac{-3}{x-y} in the next part of the solution?
    Last edited by allyourbass2212; June 25th 2009 at 08:19 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    May 2009
    Posts
    527
    \frac{3}{y - x} + \frac{x}{x - y} = \frac{-3 + x}{x - y}

    I just find it easy to remember this property:
    a - b = -(b - a)

    So the denominator of the first fraction would be:
    y - x = -(x - y)

    Anyway...
     \frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-({\color{red}-y+x})}+\frac{x}{x-y}= \frac{3}{-({\color{red}x-y})}+\frac{x}{x-y}
    You're talking about the changes in red? That's the commutative property of addition: a + b = b + a. So inside the parentheses:
    -(-y + x) = -(x + (-y)) = -(x - y). You switch the x and -y.

     =\frac{3}{-(x-y)}+\frac{x}{x-y}=\frac{-3}{x-y}+\frac{x}{x-y} in the last part of solving this expression how does the once negative denominator of =\frac{3}{-(x-y)} go to \frac{-3}{x-y} in the next part of the solution?
    You should also know that if you have a fraction that is negative, then the following are equivalent:
    -\frac{a}{b} = \frac{-a}{b} = \frac{a}{-b}

    We have to use this property because here:
    \frac{3}{{\color{red}-}(x-y)}+\frac{x}{x-y}
    the denominators are still not the same -- the first fraction has that negative sign outside the parenthesis. No problem -- the fraction property I stated above tells me that I can just move the negative to the numerator and the fraction is equivalent.
    \frac{{\color{red}-}3}{(x-y)}+\frac{x}{x-y}

    Hope that helps!


    01
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    May 2008
    Posts
    143
    Yes very much thank you so very much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. :(impossible factoring questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 1st 2010, 07:58 PM
  2. Factoring to reduce Questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: July 2nd 2009, 10:40 AM
  3. common factoring questions
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: February 1st 2009, 08:02 AM
  4. Two quick factoring questions?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 27th 2008, 11:58 PM
  5. Factoring Questions - Help!
    Posted in the Algebra Forum
    Replies: 18
    Last Post: September 8th 2007, 01:01 PM

Search Tags


/mathhelpforum @mathhelpforum