1. ## 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).

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

Sam

2. ## Re: Factoring By Grouping

Originally Posted by ArcherSam
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).

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

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

I hope this helps.

3. ## Re: Factoring By Grouping

You almost have it. Why did you quit at "?"?? Simplify that mess in the square brackets.

4. ## Re: Factoring By Grouping

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

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