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Math Help - Factoring

  1. #1
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    Unhappy Factoring

    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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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Gerard View Post
    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
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  3. #3
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    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)]
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  4. #4
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    e^(i*pi)'s Avatar
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    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)
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  5. #5
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    OHHH everything seems to have clicked in my head now thank you.
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  6. #6
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    Quote Originally Posted by Gerard View Post
    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)
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