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Math Help - Stumbling! Help!!

  1. #1
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    Stumbling! Help!!

    Need help on how to approach these specific probs.

    expand:
    (2x^2-1)^2

    Factor completely (factor common first)
    y^46y^3+9y^2

    factor/simplify lowest terms
    9x^2-1 * x^2-1
    __x-1 ___6x^2-2x

    1 - 1
    x_ x+2
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  2. #2
    Jen
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    Quote Originally Posted by dude View Post

    expand:
    (2x^2-1)^2
    On this one just re-write it as (2x^2-1)(2x^2-1) and foil it out (use the distributive law). After foiling and collecting like terms you should get 4x^4-4x^2+1
    Last edited by Jen; February 23rd 2008 at 04:30 PM.
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  3. #3
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    expand:
    (2x^2-1)^2

    =(2x^2 -1)(2x^2 -1)
    =4x^4-2x^2-2x^2+1
    =4x^4 -4x^2 +1

    Factor completely (factor common first)
    y^46y^3+9y^2

    =y^2(y^2 -6y +9)
    =y^2(y-3)(y-3)
    or y^2(y-3)^2


    factor/simplify lowest terms
    9x^2-1 * x^2-1
    ______ ______
    x-1 6x^2-2x

    =(9x^2-1)/(x-1)*(x^2-1)/(6x^2-2x)
    =(3x-1)(3x+1)/(x-1)*((x+1)x-1)/2x(3x-1)
    =(3x+1)(x+1)/2x




    1/x-1/(x+2)
    ((x+2)-x)/(x(x+2))
    2/(x^2+2x)
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  4. #4
    Jen
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    Quote Originally Posted by dude View Post

    Factor completely (factor common first)
    y^46y^3+9y^2
    On this one try to find factors that each term have in common i.e. is there some way that you can re-write each term such that they each have a similar item?

    for example I can re-write the original statement as
    y^2(y^2)-y^2(6y^1)+y^2(9) notice that if you multiply each of them back together it would make the same statement that we started with, and even more, notice that they each have a y^2 in common that we can factor out.

    so if we factor out the y^2 we get y^2(y^2-6y+9) Once again notice that if we multiply everything back out we will get the same as we started with.

    Now we can factor what we have left to y^2(y-3)(y-3) That last factoring is a little tricky, it is like undoing foiling.
    so we should finish with y^2(y-3)^2

    Hope this helps a little.
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