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Homework Help: Factoring a complex equation:

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data

    A simple way to factor the left hand side of the given equation.

    2. Relevant equations


    3. The attempt at a solution

    I would, just that i am confused on eactly how to.

    Could anyone help me simplify by factoring the left hand side of this equation in an easy and sensible manner?


    Thanks up-front.
  2. jcsd
  3. Nov 11, 2008 #2
    That's one of your common terms.
    [tex]({\frac{\frac{a^2}{b}}{b}})^2 = \frac{\frac{a^4}{b^2}}{b^2}[/tex]
  4. Nov 11, 2008 #3


  5. Nov 11, 2008 #4
    that wont work. if you multiply out something like:
    [tex](A+B) * (A+B)[/tex]
    you get
    [tex](A^2 + 2AB + B^2)[/tex]

    the expression you wanted to simplify is in the form of
    [tex](A^2 - A^2B + B^2)[/tex]
    if you had
    [tex]A = E[/tex]
    [tex]B = \frac{\frac{F^2vt}{c}}{M}[/tex]

    looking at the whole equation, couldn't you have subtracted [tex](\frac{\frac{F^4v^2t^2}{c^2}}{M^2})[/tex]
    from both sides and just end up with a simple
  6. Nov 11, 2008 #5

    You see, i am quite familiar with simple factoring of equations. I am just getting very confused with all these symbols and how to use them when i do factorize. Could you not just show me please, and put me out of my misery. My heads sore.

    As for the latter question friend, the added [tex](\frac{\frac{F^4v^2t^2}{c^2}}{M^2})[/tex] I cannot subtract.
  7. Nov 11, 2008 #6
    Why don't you just work with the equation above instead of all that mess? It's much easier to work with 2 variables, A and B rather than all that frazzle dazzle. The purpose of this forum is not to give out answers but provide hints. See if you can factor
    [tex](A^2 - A^2B + B^2)[/tex]

    You can rewrite the equation above to:
    [tex]-(A^2(B-1) - B^2)[/tex]
    That's a difference of two squares factorization.
  8. Nov 11, 2008 #7
    Agreed. That way is so much easier.

    If that is all, thanks a bunch. It was really helpful.
  9. Nov 11, 2008 #8
    (But i'll still try and solve this - I'm not off yet ;) ) Plus, i've never seen an equation of the form [tex](A^2-A^2B+B)[/tex]... i don't get out much. :)
  10. Nov 11, 2008 #9
    I need to ask so i understand this more clearly, you want me to substitute the B's of [tex](A^2+A^2B+B^2)[/tex] for [tex]\frac{\frac{F^2vt}{c}}{M}[/tex], but what about [tex]\frac{\frac{F^4v^2t^2}{c^2}}{M^2}[/tex]? Is this going to be A? What about [tex]E^2-E^2[/tex]?

    I am a terrible problem solver. There is quicker way through this, and trying to get through my thick skull as noble as it is, is not it.
  11. Nov 11, 2008 #10
    If you let
    [tex]B = \frac{\frac{F^2vt}{c}}{M}[/tex]
    [tex]B^2 = \frac{\frac{F^4v^2t^2}{c^2}}{M^2}[/tex]


    [tex]A = E[/tex]
    [tex]A^2 = E^2[/tex]
  12. Nov 11, 2008 #11
  13. Nov 11, 2008 #12
    So if you just keep in mind that A and B represents those things, once you've factored
    [tex](A^2 - A^2B + B^2)[/tex]
    You can plug in the original values of A and B into your answer to get what you wanted in the first place.
  14. Nov 11, 2008 #13
    So why is there two [tex]A^2[/tex]'s in [tex](A^2-A^2B+B^2)[/tex] - just a typo?
  15. Nov 11, 2008 #14
    [tex]A = E[/tex]
    And there are two [tex]E^2[/tex]'s in the original expression.
  16. Nov 11, 2008 #15
    Oh no, i get it now... my mistake
  17. Nov 11, 2008 #16
    Oh please can you just show me. I am trying over at this end, but none of the answers i got i don't like. Just tell me please?
  18. Nov 11, 2008 #17
    Is this what you are on about? [tex](A^2-B)(A^2+B^2)[/tex]... No?
  19. Nov 11, 2008 #18
    It's similar to that, but there's a sleight of hands because of the [tex]-A^2B[/tex] term inside. I gave you a pretty good hint [tex]-(A^2(B-1) - B^2)[/tex]... You're almost there!! =D
  20. Nov 11, 2008 #19
    Patience please, but what is the (minus one) part? Why has it been plugged in there? I need to know these things, otherwise i am lost. I assume it has something to do with subtracting one of the [tex]E^2[/tex]?
  21. Nov 11, 2008 #20
    Nope the minus sign in there is just another sleight of hands. Multiply it out and you should verify that:
    [tex]-(A^2(B-1) - B^2) = A^2 - A^2B + B^2[/tex]
  22. Nov 11, 2008 #21
    Well, i get [tex]A^22B^2=A^2-A^2B+B^2[/tex]? This aint right is it?
  23. Nov 11, 2008 #22
    Can't quite follow what you did there... What I did was basically rewrite the expression
    [tex]A^2 - A^2B + B^2[/tex]
    because I do not know how to factor that expression at that state, but I do know how to factor
    [tex](B^2 - A^2(B-1))[/tex] or [tex]-(A^2(B-1) - B^2)[/tex]

    All three expressions are equivalent to each other.
  24. Nov 11, 2008 #23
    Then i too admit i do not know how to factor the latter state, so please assist me in my teaching. Cheers,
  25. Nov 11, 2008 #24
    And take me through it too...?
  26. Nov 11, 2008 #25
    I suppose you've struggled far enough.
    Let's look at:
    [tex]-(A^2(B-1) - B^2)[/tex]
    We agree that the expression is equivalent to
    [tex](A^2 - A^2B + B^2)[/tex]
    Which is equivalent to your original complex expression that you wanted to factor.

    Disregard the minus sign and the [tex](B-1)[/tex] on the first expression, and you're left with something you might be familiar with:
    [tex](A^2 - B^2)[/tex]
    Which is the difference of squares that I mentioned about.
    We know how to factor that, right?

    Now, what to do with [tex](B-1}[/tex] that we left out? Since we're dealing with difference of squares, let's play with [tex]\sqrt{(B-1)}[/tex] and stick that to [tex]A[/tex] to get:
    [tex](A\sqrt{(B-1)} - B) (A\sqrt{(B-1)} + B)[/tex]
    If you put the minus sign back into the expression above to get
    [tex]-(A\sqrt{(B-1)} - B) (A\sqrt{(B-1)} + B)[/tex]
    And multiply it out, you should get back to the original expression:
    [tex]-(A^2(B-1) - B^2)[/tex]

    After you've verified that
    [tex]-(A\sqrt{(B-1)} - B) (A\sqrt{(B-1)} + B)[/tex]
    is indeed the factor of the first expression of this post, then all you have to do is replace A and B with its original forms.

    So there. You've basically factored that complex equation with a few sleight of hands coupled with trial and error.
    Last edited: Nov 11, 2008
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