# Factoring

• Sep 12th 2010, 11:40 AM
Gerard
Factoring
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 AM
e^(i*pi)
Quote:

Originally Posted by Gerard
How does one go about factoring

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

Please show me how to solve this.

You may combine the last two terms, just like $-7x-15x = -22x$

$2(3m+1)^2 - 22(3m+1)$

You now need to find the highest common factor of both terms. One factor is 2 because both 2 and 22 are divisible by 2
• Sep 12th 2010, 11:53 AM
Gerard
I did that but the answer on the sheet I was given is

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

While if I do it how you would I get

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

$2(3m+1)[(3m+1)-11] = 2(3m+1)(3m-10)$
• Sep 12th 2010, 12:02 PM
Gerard
OHHH everything seems to have clicked in my head now thank you.
• Sep 12th 2010, 12:37 PM
HallsofIvy
Quote:

Originally Posted by Gerard
How does one go about factoring

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

Please show me how to solve this.

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)