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Factoring out a negative

  1. Feb 10, 2017 #1
    • Member warned that the homework template must be used
    i have no idea how to use the punctuation marks for writing equations..ill do my best..

    (3a+4b)(a-3b) over b(3a+4b) multiplied by 1 over b^2(3b-a)(3b+a)

    during the cross cancelling the 3b-a would be multiplied by a negative one to switch the problem over right? (-1)(3b-a) in order to cross cancel with the (a-3b) in the top left..
    the answer was 1 over b^3(a-3b) but i dont understand the whole negative sign thing..isnt (3b-a) the same as (-a+3b)? so factoring out the negative would make it (a-3b) and it cancels with the top left fraction but then i also have the (b) (b^2)(3b+a) on the bottom so how do they get b^3(a-3b)? i get the b^3 part ..i thought maybe the negative one i factored out in the (3b-a) would have also applied to the (3b+a) but then wouldnt it be (-3b-a) ..im so confussed on factoring out negatives..
     
  2. jcsd
  3. Feb 11, 2017 #2

    fresh_42

    Staff: Mentor

    Sure there isn't a typo? What happened to ##(3b+a)##?
    see above
    I don't really see where the ##-1## has gone to, neither in the quoted answer nor in the confusion of your reasoning.

    To write formulas, you should read
    https://www.physicsforums.com/help/latexhelp/

    It's not that difficult but makes reading a lot easier. I think you have all you need to simplify the quotient. Just keep track of your parts. A simple trick to see, whether a step is write or wrong, is to plug in some small numbers for ##a## and ##b##, like ##\pm 1## or ##\pm 2##.
     
  4. Feb 11, 2017 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Was it the problem?
    [tex]\frac{(3a+4b)(a-3b)}{b(3a+4b)}\frac{1}{b^2(3b-a)(3b+a)}[/tex]
    [tex]=\frac{(3a+4b)(a-3b)}{b(3a+4b)}\frac{1}{-b^2(a-3b)(3b+a)}[/tex]

    You cancelled out a-3b, and you got a (-1) factor in the denominator. Dividing by -1 makes the sign of the whole fraction negative, so it is
    [tex]-\left(\frac{(3a+4b)}{b(3a+4b)}\frac{1}{b^2(3b+a)}\right)[/tex]after further simplification, you get the desired formula.

    The minus sign is the same as a factor of (-1).
    Remember the rules when calculating with factors. [tex]\frac{a}{bc}=\frac{1}{b}\frac{a}{c}[/tex]
    If b = -1,
    [tex]\frac{a}{-c}=\frac{1}{-1}\frac{a}{c}[/tex]
    and you know that [tex]\frac{1}{-1}= -1[/tex]
    so [tex]\frac{a}{-c}=-\frac{a}{c}[/tex]
     
  5. Feb 11, 2017 #4
    #3 ..writing it out not that hard? looking at that page just made my eyes cross and my brain short circuit..ugh forget it..
     
  6. Feb 11, 2017 #5

    Ray Vickson

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    Science Advisor
    Homework Helper

    Can you write (3a+4b)(a-3b)/( b(3a+4b))? Is that easier than writing \frac{(3a+4b)(s-3b)}{b(3a+4)}? Basically, that's all there is to it!

    When you write x = \frac{a}{b} in LaTeX, you get ##x = \frac{a}{b}## after you enclose everything between two # symbols, like this: # # your stuff # # (remove spaces between the two # signs at the start and the end). Using # delimiters produces an "in-line" formula. If you want a "displayed" formula/equation, like this
    $$ x =\frac{a}{b},$$
    you should replace the # symbols by $ signs, so write $ $ your stuff $ $ (with no spaces between the two $ signs at the start and the end).
     
    Last edited: Feb 11, 2017
  7. Feb 12, 2017 #6
    no idea what that means...the first thing u wrote made sense, the rest way over my head..this site wont let me post pictures either otherwise id just upload a pic of the problem
     
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