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Identitys Are SO CONFUSING

  1. Oct 11, 2004 #1
    Hey guys,

    Having trouble gettign my head around these indentities!
    So confusing so does anyone have a website which goes through the explanations systematically.

    My text books touches on it but the explanation doesn't cover all the variations.
    and the "letts AS Revisions guide" doesn't cover it at all which is rather annoying as i waitied for it and thought it would!

    Please help,

    I need a grade A in maths AS and its just holding me back!
  2. jcsd
  3. Oct 11, 2004 #2


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    Which identities would those be?
  4. Oct 11, 2004 #3
    "olving the square" ones. And others such as
    x^3 + ax^2 -3x + b =(the three line = Symbol) (x -c)(x - 4) (x+ 5)
    i understand the way to solve some using the "values" method.

    But the "coefficents" or "Comparing coefficents" method is confusing.
  5. Oct 11, 2004 #4


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    I don't know what you mean by "solving the square."

    Actually, comparing coefficients is probably the simplest way for determining the values but, conceptually, it might be easier to try something like the following. Replace x with specific values and obtain equations for the unknowns! In this case, good ones are x = 4 and x = -5 which give simple equations for a and b that you can easily solve. A third value, given a and b, would be x = 0 in order to obtain c. Oftentimes, trying another method will give you insights and perhaps even an appreciation of the "standard" techniques.

    One of the really nice things about math is that there are often many ways of obtaining solutions to problems though math courses sometimes focus on a single approach.
  6. Oct 11, 2004 #5


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    Identities are always true, that's the 1st thing you need to remember. It doesn't matter what value of x or theta or whatever they are always true.

    Simple identity:

    [tex]1 + 1 \equiv 2[/tex]

    That is always true, it does not depends on a specific value of x. So if:

    [tex]A + 1 \equiv 2[/tex]

    Where A is some constant, then we know A = 1. Carrying on to something more complex:

    [tex]x + x + 1 + 1 \equiv 2x + 2[/tex]

    Notice that x doesn't equal anything particular, this is just always true. So taking another example:

    [tex]x + 2x + x + 1 + 76 \equiv Ax + B[/tex]

    We can easily work out that B = 77 and A = 4 for any value of x. So key thing is, always make as simple as possible and only compare the things in front of the same number of x's. So following on taking your example:

    [tex]x^3 + ax^2 - 3x + b \equiv (x-c)(x-4)(x+5)[/tex]

    [tex]x^3 + ax^2 - 3x + b \equiv x^3 + 5x^2 - 4x^2 - cx^2 + 4cx - 5cx - 20x + 20c[/tex]

    [tex]x^3 + ax^2 - 3x + b \equiv x^3 + (1-c)x^2 + (-20-c)x + 20c[/tex]

    Now the 1st thing to notice is that we have an easy relationship look at just the bit in front of x. We have a [itex]-3[/itex] in front of the one on the left hand side and a [itex]-20-c[/itex], now as they are always equal no matter what value of x that must mean that: [itex]-3 = -20-c[/itex] so we can say that:

    [tex]3 = 20 + c[/tex]

    [tex]c = -17[/tex]

    Now we have the c = -17 lets look at the bit in front of the [itex]x^2[/itex] term. We have that 1 - c is the same as a, now it is easy to get from this that a = 18. Finally we have 20c = b as both of these are on their own without an x, so this gives us that b = -340.

    Hope that helps.
  7. Oct 12, 2004 #6
    That kind of helps and i also saw the book get (1 - a) from ax^2 etc. etc.

    How did you get that? Are you simply expanding it so that you have a number to work with and because there is no coefficient for A it must be 1?

    (Note: im talking about the +(1 - c)x^2 )

    Also, Expanding three brackets at a time got me bit confused. I normally use the "FOIL" technique so i would have done:

    First ones (x X x X x) = x^3
    Outsides: x X 5 = 5x
    Inside . . . . Then realise its all messed up!

    How should i do those () () () ? As i say, I usually only do (x + x) (x - x) for example.

    Once again there are to many unexplained steps :(

    I don't see how you got from

    https://www.physicsforums.com/latex_images/33/339434-5.png [Broken]
    https://www.physicsforums.com/latex_images/33/339434-6.png [Broken]

    I see you kept the x^3, 20c then everything else seems to have somehow changed :S

    I need a website explaining all the possible techniques, I found Values (i.e. substituing x = 1, x = 2 etc) easy. But then it doesn't seem to work for all of them, or does it?

    This CoEfficent technique seems to take huge steps and i dont see how they can go from one thing to the other.

    would it be possible to go from:


    to group similar items and get:

    so i would then get:x^3+x^2-cx^2-cx-20x+20c

    Then im all stuck! and don't know what to do next.

    I seem to be treating them like equations and thinking "well i have x^3 on both sides so i can cancel that out". But i know i can't.

    Hence i really need a deep explanation .

    Look forward to response,
    Last edited by a moderator: May 1, 2017
  8. Oct 12, 2004 #7


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    As my Gran always used to tell me when I was 1st learning algebra (she was a maths teacher), collect like terms!

    If you have [itex]x + x^2 + x[/itex] this is the same as: [itex]x^2 + 2x[/itex]. Another example [itex]cx + 2x^3 - 3x[/itex] is the same as: [itex]2x^3 + (c - 3)x[/itex].

    This is all I have done, for example looking purely at the [itex]x^2[/itex] term, I had [itex]5x^2 - 4x^2 -cx^2[/itex] is the same as: [itex]x^2 - cx^2[/itex] taking x^2 as a common factor: [itex](1 - c)x^2[/itex].

    As for dealing with the product of 3 linear equations. Well if your confident enough with quadratic equations you can take this:


    And rearrange it like so:


    Now just expand the [itex](cx+d)(ex+f)[/itex] then finally expand whatever is left that needs expanding and collect like terms.

    Hope that helps :smile:
  9. Oct 14, 2004 #8
    Thank zurtex, thats made me realise i REALLY REALLY need to touch up on my expanding/Factoring alot!

    Got three basic rules now:
    Collect like terms
    Factorise where possible
    Look at it again and follow the above steps :P

    Ok i'm off to relearn factorising and expanding as i have clearly forgotten alot of it!
  10. Oct 14, 2004 #9
    Definatly need to practice... this has puzzled me!

    (x^2 - a/x^2) (x^2 - a/x^2)

    I know the answer is:


    Problem is when i tried it...

    i got it all wrong!

    i normally use the highschool "FOIL" technique...

    But i can't use it on that!

    I get things like x^4 X a/x^2

    gets messy :)
    Last edited: Oct 14, 2004
  11. Oct 14, 2004 #10


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    [tex]x^4 * \frac{a}{x^2} = x^4 * \frac{1}{x^2} * a = \frac{x^4}{x^2} * a = ax^2[/tex]

    Anyway, looking at this: (x^2 - a/x^2) (x^2 - a/x^2). Take it from this approch:

    [tex](y - z)^2 = (y - z)(y - z) = y^2 - 2yz + z^2[/tex]

    Now let y = x^2 and z = a/x^2
  12. Oct 14, 2004 #11
    yeh i did it like that but it got sooooo messy and screwed up!

    I mean what does: x^2 multiplyed a/x^2 =?

    Really confused lol! Need to learn this expanding stuff with / involved!

    never done them much before and when i did, i got them all wrong

    Please post your method fully explained if you have time.
  13. Oct 14, 2004 #12

    matt grime

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    it equals a. (ie neither of your guesses, sorry)

    a/x^2 is the same as ax^{-2} and whenever you multiply x^{p} by x^{q} the answer is x^{p+q} for all p and q.

    so here you have p=2 and q=-2, so 2+-2 = 0 and x^0=1.
  14. Oct 14, 2004 #13
    my teacher says its:

    (x^2 +a/x^2)(x^2 +a/x^2) + b

    = x^4 -2a +a^2/x^4 +b
  15. Oct 14, 2004 #14

    matt grime

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    excepting the minus sign error(s), yes. (which is what foil would give you too. but don't use foil: it is unnecessary).
  16. Oct 14, 2004 #15
    but i need to know how to get there lol :)

    anyone got good website with expanations on / In expanding!

    I have never seen Divisions in () ()'s

    So need to get my head around them

  17. Oct 14, 2004 #16

    matt grime

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    the division is a red herring!

    As some one has pointed out (u+v)^2 = u^2+2uv+v^2 is true what ever u and v are. after you've done that you simplify the terms.

    and if we need to rehash it:

    (u+v)(u+v) = u(u+v) + v(u+v) = u^2+uv+vu+v^2, rearranging gives us the answer we want.
  18. Oct 15, 2004 #17
    which minus sign errors?

    when i try this i end up getting:

    x^4 -2a/x^8 -a^2/x^4

    Must have gone wrong with the +/- somewhere because my tutor has 2a over nothing.

    But i have 2a/x^8.

    So when simplifying i assume i should of had -a/-x^4 and -a/+x^4 therefore cancelling eachother out.

    I seem to have forgotten the basics on expanding and factorising :(
  19. Oct 15, 2004 #18

    matt grime

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    you seem to have forgotten the basics on exponents:

    you have x^2 times a/x^2 which simplifies to a, not a/x^8.

    your post three back contains a minus sign error. read it again. notice you hav(x^2+a/x^2) in the brackets, not minus as you originally wrote.
  20. Oct 15, 2004 #19
    yeh i definatly have. Know any websites which have good explanation of this stuff?

    I need to revise it before continuing and im really far behind in my studys! at this rate ill of read everything once and will have 0 Revision time :(

    Do you have a webpage or any ebooks with this explained?

    Can't find my old GCSE revision stuff!

    and the sites i have been on do not have any explanations of this stuff (specially with Division involved).
  21. Oct 15, 2004 #20
    if x^2 X -a/x^2 = a

    then how did my tutor get

    -a/x^2 X a/x^2 = -a^2/x^4 ?

    Wouldnt it be the same rule meaning -a/x^2 X -a/x^2 = +a^2

    I got +a^2/x^4

    i guess it wouldnt as its both DIVIDED BY x^2 rather than
    x^2 X a/x^2

    Hmm, clearly need to revise this but can't find anywhere to do it!
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