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Math Help - factoring hmwk...please help?

  1. #1
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    factoring hmwk...please help?

    i am terrible at factoring and i would like some much needed help with these!

    a) 36(2x - y)^2 - 25(u - 2y)^2
    b) x^2 - y^2 + z^2 - 2xz
    c) x^2 + 2 + 1/x^2
    d) p^2 - 2p + 1 - y^2 - 2yz - z^2

    thanks!!!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by checkmarks View Post
    c) x^2 + 2 + 1/x^2
    d) p^2 - 2p + 1 - y^2 - 2yz - z^2
    I can help you with these, anyway:
    c) x^2 + 2 + \frac{1}{x^2}

    If this were a^2 + 2ab + b^2 you would have no trouble factoring it (I presume).

    Let a = x and b = \frac{1}{x}

    Then your expression becomes:
    a^2 + 2ab + b^2 = (a + b)^2 = \left ( x + \frac{1}{x} \right ) ^2

    This is a useful little trick to remember. You'll see this kind of factoring every now and again.

    d)
    p^2 - 2p + 1 - y^2 - 2yz - z^2

    (p^2 - 2p + 1) - (y^2 + 2yz + z^2)

    (p - 1)^2 - (y + z)^2

    Nothing more you can do here.

    -Dan
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  3. #3
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    Quote Originally Posted by checkmarks View Post
    a) 36(2x - y)^2 - 25(u - 2y)^2
    b) x^2 - y^2 + z^2 - 2xz
    The first one can be solved using the identity a^2-b^2=(a+b)(a-b)

    The second one it's similar, look for a perfect square, then factorice.

    Quote Originally Posted by topsquark View Post
    d)
    p^2 - 2p + 1 - y^2 - 2yz - z^2

    (p^2 - 2p + 1) - (y^2 + 2yz + z^2)

    (p - 1)^2 - (y + z)^2

    Nothing more you can do here.
    You sure?
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    The first one can be solved using the identity a^2-b^2=(a+b)(a-b)

    The second one it's similar, look for a perfect square, then factorice.
    i figure that but i dont know what to do with the 36 and 25
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  5. #5
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    Quote Originally Posted by checkmarks View Post
    a) 36(2x - y)^2 - 25(u - 2y)^2
    36(2x-y)^2-25(u-2y)^2=[6(2x-y)]^2-[5(u-2y)]^2

    Now apply a^2-b^2=(a+b)(a-b)
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    The first one can be solved using the identity a^2-b^2=(a+b)(a-b)

    The second one it's similar, look for a perfect square, then factorice.


    You sure?
    (shaking his head in despair)

    So just how many times have you had to correct me today, Krizalid?

    -Dan
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  7. #7
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    I hope you don't get mad.

    There're just details, it's nothing to worry about
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    I hope you don't get mad.

    There're just details, it's nothing to worry about
    I only get mad at myself. I don't mind making the occasional mistake, because everyone does at some point and because there are skilled workers here who will catch them.

    What I get frustrated about is when I keep making mistakes over and over and over... That means I'm simply working too fast and that doesn't really help the people who have asked for it.

    (Not to mention I have nearly as god-like an ego as ThePerfectHacker. Don't tell him that, though, because he might feel that there isn't enough Hilbert space here for the two of us! )

    -Dan
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