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April 16th, 2015, 10:57 PM   #1
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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 01:02 AM.
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April 17th, 2015, 01:13 AM   #2
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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)
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April 17th, 2015, 02:09 AM   #3
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Quote:
Originally Posted by skipjack View Post
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
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