How does one go about factoring

2(3m+1)^2-7(3m+1)-15(3m+1)

Please show me how to solve this.

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- Sep 12th 2010, 11:40 AMGerardFactoring
How does one go about factoring

2(3m+1)^2-7(3m+1)-15(3m+1)

Please show me how to solve this. - Sep 12th 2010, 11:44 AMe^(i*pi)
- Sep 12th 2010, 11:53 AMGerard
I did that but the answer on the sheet I was given is

$\displaystyle 2(3m+1)(3 m-10)$

While if I do it how you would I get

$\displaystyle 2[(3m+1)-11(3m+1)]$ - Sep 12th 2010, 11:58 AMe^(i*pi)
$\displaystyle m+1$ is also a factor of both terms so that can be taken out too

$\displaystyle 2(3m+1)[(3m+1)-11] = 2(3m+1)(3m-10) $ - Sep 12th 2010, 12:02 PMGerard
OHHH everything seems to have clicked in my head now thank you.

- Sep 12th 2010, 12:37 PMHallsofIvy
The first thing I see is that there is a "3m+1" in each term. Taking that out leaves (3m+1)(2(3m+1)- 7- 15). Now you can combine -7- 15= -22 as e^(i*pi) suggests as well as multipy out 2(3m+1) to get (3m+ 1)(6m+ 2- 22)= (3m+1)(6m- 20). If you really want to you could observe that both 6m and 20 are divisible by 2 so you can factor a little further: 2(3m+1)(3m- 10)