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 April 16th, 2015, 11:57 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: $\displaystyle 12x^2-21x-6$ I have understood that I need to factor out the greatest common factor which is 3. So I have $\displaystyle 3(4x^2-7x-2)$ 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 $\displaystyle 3[(4x^2-8x+x-2)]$ If I factor the first group $\displaystyle (4x^2-8x)$ I get $\displaystyle 3[x(4X+1)]$ But the second group $\displaystyle (4x^2-8x)$ 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 02:02 AM.
 April 17th, 2015, 02:13 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,422 Thanks: 1462 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) Thanks from mikeinitaly
April 17th, 2015, 03:09 AM   #3
Senior Member

Joined: Oct 2012
From: Live and work in Saudi but from Italy (British national)

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Math Focus: Hoping to do something in the future.
Quote:
 Originally Posted by skipjack 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)
OK I've got it
Thanks

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