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Math Help - Factoring By Grouping

  1. #1
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    Talking Factoring By Grouping

    Hey

    I'm having trouble factoring by grouping. I rearrange the equation then I factor out the common factors and now i'm lost . Help! LOL(funny smiley).

    \begin{array}{l} 2(x + 3)(x - 2){}^3 + (x + 3){}^2 \circ 3(x - 2){}^2\\ 2(x + 3)(x - 2){}^3 + 3(x - 2){}^2(x + 3){}^2\\ (x + 3)(x - 2){}^2[2(x - 2) + 3(x + 3)]\\ ?\\ Answer = 5(x + 3)(x - 2){}^2(x + 1) \end{array}

    Sam
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Re: Factoring By Grouping

    Quote Originally Posted by ArcherSam View Post
    Hey

    I'm having trouble factoring by grouping. I rearrange the equation then I factor out the common factors and now i'm lost . Help! LOL(funny smiley).

    \begin{array}{l} 2(x + 3)(x - 2){}^3 + (x + 3){}^2 \circ 3(x - 2){}^2\\ 2(x + 3)(x - 2){}^3 + 3(x - 2){}^2(x + 3){}^2\\ (x + 3)(x - 2){}^2[2(x - 2) + 3(x + 3)]\\ ?\\ Answer = 5(x + 3)(x - 2){}^2(x + 1) \end{array}

    Sam
    You are very close. You already have the (x+3)(x-2)^2 part -- you just need to simplify the 2(x-2)+3(x+3) bit; first distribute and then combine like terms. The answer will become apparent once you do so.

    I hope this helps.
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  3. #3
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    Re: Factoring By Grouping

    You almost have it. Why did you quit at "?"?? Simplify that mess in the square brackets.
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  4. #4
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    Re: Factoring By Grouping

    I feel so ashamed. It was that easy. LOL.

    \begin{array}{l} 2(x + 3){(x - 2)^3} + {(x + 3)^2}^\circ 3{(x - 2)^2}\\ 2(x + 3){(x - 2)^3} + 3{(x - 2)^2}{(x + 3)^2}\\ (x + 3){(x - 2)^2}[2(x - 2) + 3(x + 3)]\\ (x + 3)(x - 2){}^2[2x - 4 + 3x + 9]\\ (x + 3)(x - 2){}^2(5x + 5)\\ (x + 3)(x - 2){}^25(x + 1)\\ 5(x + 3)(x - 2){}^2(x + 1)\\ Answer = 5(x + 3){(x - 2)^2}(x + 1) \end{array}
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