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April 16th, 2015, 10:57 PM  #1 
Senior Member Joined: Oct 2012 From: Live and work in Saudi but from Italy (British national) Posts: 103 Thanks: 2 Math Focus: Hoping to do something in the future.  Factoring by grouping
Hello all. Please forgive me for repeating myself. I am a beginner in maths and I am learning this in my free time because I enjoy it. I really want to understand this stuff and I have watched numerous videos and read examples, but I can't understand where I am going wrong here. I have this trinomial and I am trying to factor it by grouping: $\displaystyle 12x^221x6$ I have understood that I need to factor out the greatest common factor which is 3. So I have $\displaystyle 3(4x^27x2)$ Now I need to expand this to a 4 term polynomial so I can factor by grouping. Thus 4x(2)=8 Now I need two integers which give a product of 8 and a sum of 7 thus 1, 8 Expanding the middle term gives $\displaystyle 3[(4x^28x+x2)]$ If I factor the first group $\displaystyle (4x^28x)$ I get $\displaystyle 3[x(4X+1)]$ But the second group $\displaystyle (4x^28x)$ doesn't make sense because I get $\displaystyle 2(4x+3)$. Putting it all together, I get $\displaystyle 3[x(4x+1)2(4x+3)]$ Help! M Last edited by skipjack; April 17th, 2015 at 01:02 AM. 
April 17th, 2015, 01:13 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1385 
3(4x²  8x + x  2) = 3[4x(x  2) + 1(x  2)] = 3(4x + 1)(x  2) 3(4x² + x  8x  2) = 3[x(4x + 1)  2(4x + 1)] = 3(x  2)(4x + 1) 
April 17th, 2015, 02:09 AM  #3 
Senior Member Joined: Oct 2012 From: Live and work in Saudi but from Italy (British national) Posts: 103 Thanks: 2 Math Focus: Hoping to do something in the future.  

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