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Expanding parenthesis when a negative is involved

  1. Oct 27, 2012 #1
    Im completing Engineering Maths cover to cover in an attempt to get more familiar with maths as I finished my education many years ago without really understanding many basic maths concepts. This problem is at the back of the introduction to algebra.

    (x-2y)^2 - (2x - y)^2

    Now I can expand fairly easily:

    x(x-2y) - 2y(x-2y) - (2x)(2x-y) - (-y)(2x-y)

    x^2 - 2yx - 2yx + 4y^2 - 4x^2 ... Hmm

    I know the rest as its in the book but I don't know the correct method for determining:
    - (-y)(2x-y)

    because of the negative at the front. without it, its easy:
    -y * 2x = -2xy
    -y * -y = y^2

    thus:
    -2xy + y^2

    but then you have ... - -2xy + y^2

    minus minus???

    What I dont understand is how the negative before the brackets affects the result. When expanding is it:
    - (-y)(2x-y)
    -y * 2x = -2xy * -1 = 2xy
    -y * -y = y^2 * -1 = -y^2

    or maybe the - term belongs to all the last bit?

    - (-2xy + y^2)
    so its
    -1 * (-2xy + y^2)
    = +2xy - y^2

    If someone could explain the rule dictated to expand when theres a negative, I would be grateful as I cant find an example or comment about how to think of this.

    Of course theres the other FILO way (a+b)(c+d) = ac+bc+bd+ad

    but thats just confusing me more WRT expanding the two elements of (x-2y)^2 - (2x - y)^2 because of the negative.

    My maths is riddled with these inconsistencies where I just used to guess without understanding whats missing or what rule to follow.

    (apologies for the stupid question, thanks for any help)
     
  2. jcsd
  3. Oct 27, 2012 #2
    This is very elementary stuff... a high school teacher could probably explain it better. You can always replace a - by a multiplication with (-1). So [tex]- (-y)(2x-y) = (-1)\cdot (-y) \cdot (2x-y) [/tex] Now there are many ways to calculate the results. The easiest is this one: [tex](-1)\cdot ((-1) \cdot y) \cdot (2x - y) = (-1)\cdot (-1) \cdot y \cdot (2x - y)= y \cdot (2x - y) = 2xy - y^2[/tex] because [tex](-1)\cdot(-1)=1[/tex] Btw [tex]a - b = a + (-1)\cdot b[/tex] and your equation would usually be solved using the binomial theorem...
     
  4. Oct 27, 2012 #3

    haruspex

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    Looks to me that you got the right answer by both methods above.
    You can think of -n as +(-1)*n. So - (-y)(2x-y) = +(-1) (-y)(2x-y). Multiplying the first two terms together gives +y(2x-y). Or multiplying the last two terms first gives +(-1)*(-2xy+y^2) = 2xy-y^2.
     
  5. Oct 27, 2012 #4

    arildno

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    The simplest, and safest way is to introduce a new parenthesis between the expression to be expanded, and then solve that paranthesis later. You can ALWAYS place as many parentheses you want about a single term.

    So:
    -(2x-y)^2=-((2x-y)^2). Now, internally, there is no dangerous minus sign, so you can proceed
    -((2x-y)^2)=-(2x(2x-y)-y(2x-y))=
    -(4x^2-2xy-2xy+y^2)=-(4x^2-4xy+y^2)=-4x^2+4xy-y^2
     
  6. Oct 27, 2012 #5
    Annoyingly Id just typed out a long response and the session timed out when I went to post it :grumpy:

    Anyway,

    I managed to factorize
    (x-2y)^2-(2x-y)^2
    to
    3(y^2-x^2)
    via, as you posted introducing (-1), more parenthesis.

    i.e.
    (x)(x-2y)(-2y)(x-2y)(-1)(2x)(2x-y)(-1)(-y)(2x-y)

    The last time I was around high school teachers was 20 years ago and for whatever reason, I didnt get it then ;) Seems elementary now though.

    Thanks all :)

    Just looking at the binomial theorem.
     
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