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Thread: Factor completely

  1. #1
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    Factor completely

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

    $\displaystyle 2n^2+4ng+2g^2+n+g-3$

    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?

    $\displaystyle r^2(1+h)(1-h)+4(rh-1)$
    $\displaystyle (r-rh+2)(r+rh-2)$


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

    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
    $\displaystyle (r-rh+2)(r+rh-2)$



    $\displaystyle (n+g-1)(2n+2g+3)$



    it didn't take me that long! now try to discover what i did!
    i'm really stucked here,
    1. $\displaystyle r^2(1-h^2)+4(rh-1)$
    then i got $\displaystyle r^2-r^2h^2+4rh-4$
    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. $\displaystyle r^2(1-h^2)+4(rh-1)$
    then i got $\displaystyle r^2-r^2h^2+4rh-4$
    i don't know how to factor this
    $\displaystyle r^2(1-h^2)+4(rh-1)=r^2-(r^2h^2 - 4rh + 4)=r^2 - (rh -2)^2.$ now use the identity: $\displaystyle a^2-b^2=(a-b)(a+b).$
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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 $\displaystyle 2n^2+4ng+2g^2+n+g-3$??
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  6. #6
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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 $\displaystyle 2n^2+4ng+2g^2+n+g-3$??
    Hi princess_21,

    Here's how this one breaks out:

    $\displaystyle 2n^2+4ng+2g^2+n+g-3$

    Factor out 2 in the 1st 3 terms:

    $\displaystyle 2(n^2+2ng+g^2)+n+g-3$

    The grouped trinomial above can be factored:

    $\displaystyle 2(n+2)^2+(n+g)-3$

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

    $\displaystyle 2u^2+u-3$

    This factors into:

    $\displaystyle (2u+3)(u-1)$

    Now, back substitute:

    $\displaystyle (2(n+g)+3)(n+g-1)$

    $\displaystyle (2n+2g+3)(n+g-1)$
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