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Homework Help: Reviewing pre factoring

  1. Nov 22, 2007 #1
    [SOLVED] reviewing pre... factoring

    1. The problem statement, all variables and given/known data

    I bought a Schaum's with Precalculus questions; figured I'd review my pre. I'm not as rusty as I thought I'd be... but I'm screwing up this question for some reason:


    3. The attempt at a solution




    ?? I'm doing something wrong here.
  2. jcsd
  3. Nov 23, 2007 #2

    Gib Z

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    Complete the square: [tex](4x^2 + y^2) = 16x^4 + 8x^2y^2 + y^4[/tex].

    So now express what you have as a difference between the perfect square, and another number =]
  4. Nov 23, 2007 #3
    how did you get [tex](4x^2 + y^2)[/tex] from [tex](2x+y)^{2}[/tex]? wouldn't it be [tex](4x^2 + y^2)^{2}[/tex]? ... I'm guessing it's a typo, or else I'm really lost :bugeye: :rofl:

    ok, [tex]16x^4 + 8x^2y^2 + y^4[/tex] leaves me with [tex](4x^{2}+y^{2})^{2}-3xy[/tex] ... but wouldn't that [tex]3xy[/tex] have to be a [tex]3xy^{2}[/tex] for me to be able to do a difference of a square? ... right now it's an [tex]a^{2}-b[/tex]

    EDIT: post #4

    ugh, things like this frustrate me. I'll be doing just fine, and then a simple question like this comes along that I get all wrong... I wish I'd taken math in highschool :grumpy:
    Last edited: Nov 23, 2007
  5. Nov 23, 2007 #4
    wait... I square all of [tex]4x^{2}+4xy+y^{2}-3xy[/tex] and get



    right? ... you know, maybe the doctor's right and I do need Ritalin after all :rolleyes:.
    Last edited: Nov 23, 2007
  6. Nov 23, 2007 #5

    Gib Z

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    Well your first mistake is wrongly reducing [tex]16x^{4}-x^{2}y^{2}+y^{4}[/tex] to [tex]4x^{2}-xy+y^{2}[/tex]. What Im sure you meant was that [tex]16x^{4}-x^{2}y^{2}+y^{4} =4u^{2}-uv+v^{2} [/tex] where u= 4x^2 and v=y^2.

    Instead, from post 2, we can see what you have is [tex](4x^{2}+y^{2})^{2}-9xy^{2}[/tex], which you did factor properly =]
  7. Nov 23, 2007 #6
    o boy :blushing: there we go. thanks :biggrin:
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