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Math Help - Factor completely

  1. #1
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    Factor completely

    can i factor these completely?
    r^2(1+h)(1-h)+4(rh-1)

    2n^2+4ng+2g^2+n+g-3

    thanks for your time.
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  2. #2
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    Quote Originally Posted by princess_21 View Post

    can i factor these completely?

    r^2(1+h)(1-h)+4(rh-1)
    (r-rh+2)(r+rh-2)


    2n^2+4ng+2g^2+n+g-3
    (n+g-1)(2n+2g+3)

    thanks for your time.
    it didn't take me that long! now try to discover what i did!
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    (r-rh+2)(r+rh-2)



    (n+g-1)(2n+2g+3)



    it didn't take me that long! now try to discover what i did!
    i'm really stucked here,
    1. r^2(1-h^2)+4(rh-1)
    then i got r^2-r^2h^2+4rh-4
    i don't know how to factor this
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  4. #4
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    Quote Originally Posted by princess_21 View Post
    i'm really stucked here,
    1. r^2(1-h^2)+4(rh-1)
    then i got r^2-r^2h^2+4rh-4
    i don't know how to factor this
    r^2(1-h^2)+4(rh-1)=r^2-(r^2h^2 - 4rh + 4)=r^2 - (rh -2)^2. now use the identity: a^2-b^2=(a-b)(a+b).
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  5. #5
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    Talking

    oh, yes i got it. thank you very much!

    but how did you get the factor of 2n^2+4ng+2g^2+n+g-3??
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  6. #6
    A riddle wrapped in an enigma
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    Quote Originally Posted by princess_21 View Post
    oh, yes i got it. thank you very much!

    but how did you get the factor of 2n^2+4ng+2g^2+n+g-3??
    Hi princess_21,

    Here's how this one breaks out:

    2n^2+4ng+2g^2+n+g-3

    Factor out 2 in the 1st 3 terms:

    2(n^2+2ng+g^2)+n+g-3

    The grouped trinomial above can be factored:

    2(n+2)^2+(n+g)-3

    Now, to make things super simple, let u = (n-2).

    2u^2+u-3

    This factors into:

    (2u+3)(u-1)

    Now, back substitute:

    (2(n+g)+3)(n+g-1)

    (2n+2g+3)(n+g-1)
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