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Math Help - More Factoring

  1. #1
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    More Factoring

    I'm having some more trouble: Please help me, a little is very appreciated and a lot of help would be a God-send. Thanks, I love this community and I wish I would have found this place many years ago.
    Problem 11:
    a^2(b+c)(2b+c)+a(b+c)
    I got this far: (b+c)[a^2(2b+c)+a]
    Same as problem 12. Doesn't come out with same answer when I plug in values to check.

    Problem: 12:
    t^3(2r+s)(3r+2s)+t^2(3r+2s)
    I got this far: (3r+2s)[t^3(2r+s)+t^2]
    If I factor out t, I can check to see if it works. by doing t=5, r=4, s=3 and I get the same answer as the problem originally. However, if I factor out t^2 it makes the answer different...any reason or explanation why?

    Problem 13:
    (2a+1)(a+3)-(2a+1)(a-3)
    I got this far: (2a+1)[(a+3)-(a-3)]
    I know this is a difference of squares, but what do I do with it?

    Problem 15: 3a(5a+2b)(3a-5b)-9(a^2)(5a+2b)(4a-b)
    I got this far: (5a+2b)[3a(3a-5b)-9(a^2)(4a-b)]
    No idea where to go with this one.

    and Problem 16: 6m(3n-2m)(n+m)+8(m^2)(3n-2m)(n+m)]
    I got this far: 3n-2m[(6m)(n+m)+8(m^2)(n+m)]
    And tried : n+m(3n-2m)[6m+8(m^2)]
    Same as before, no idea where to go either.

    Any help would be appreciated.
    Thanks
    Last edited by andy011789; September 30th 2007 at 05:19 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by andy011789 View Post
    I'm having some more trouble: Please help me, a little is very appreciated and a lot of help would be a God-send. Thanks, I love this community and I wish I would have found this place many years ago.
    Problem 11:
    a^2(b+c)(2b+c)+a(b+c)
    I got this far: (b+c)[a^2(2b+c)+a]
    Same as problem 12. Doesn't come out with same answer when I plug in values to check.
    a^2(b + c)(2b + c) + a(b + c)

    Pull the a(b + c) out from both terms:
    a(b + c)[a(2b + c) + 1] = a(b + c)(2ab + ac + 1)

    Quote Originally Posted by andy011789 View Post
    Problem: 12:
    t^3(2r+s)(3r+2s)+t^2(3r+2s)
    I got this far: (3r+2s)[t^3(2r+s)+t^2]
    If I factor out t, I can check to see if it works. by doing t=5, r=4, s=3 and I get the same answer as the problem originally. However, if I factor out t^2 it makes the answer different...any reason or explanation why?
    The odds are you did something wrong with your simplification. Since you didn't show us your work, I can only speculate.

    t^3(2r + s)(3r + 2s) + t^2(3r + 2s)

    You can factor a t^2(3r + 2s):
    t^2(3r + 2s)[t(2r + s) + 1] = t^2(3r + 2s)(2rt + st + 1)

    Quote Originally Posted by andy011789 View Post
    Problem 13:
    (2a+1)(a+3)-(2a+1)(a-3)
    I got this far: (2a+1)[(a+3)-(a-3)]
    I know this is a difference of squares, but what do I do with it?
    Difference of what squares? I don't see any.

    You factored correctly, now simplify:
    (2a+1)[(a+3)-(a-3)] = (2a + 1)[a + 3 - a + 3] = 6(2a + 1)

    Quote Originally Posted by andy011789 View Post
    Problem 15: 3a(5a+2b)(3a-5b)-9(a^2)(5a+2b)(4a-b)
    I got this far: (5a+2b)[3a(3a-5b)-9(a^2)(4a-b)]
    No idea where to go with this one.
    3a(5a+2b)(3a-5b)-9(a^2)(5a+2b)(4a-b)

    You can factor a 3a(5a + 2b) from each term:
    3a(5a + 2b)[(3a - 5b) - 3a(4a - b)]

    Now simplify.

    Quote Originally Posted by andy011789 View Post
    Problem 16: 6m(3n-2m)(n+m)+8(m^2)(3n-2m)(n+m)]
    I got this far: 3n-2m[(6m)(n+m)+8(m^2)(n+m)]
    And tried : n+m(3n-2m)[6m+8(m^2)]
    Same as before, no idea where to go either.
    Remember the parenthesis!!

    6m(3n-2m)(n+m)+8(m^2)(3n-2m)(n+m)

    Again you can factor more than you did: you can take a 2m(3n - 2m)(n + m) out of both terms.

    2m(3n - 2m)(n + m)[3 + 4m]

    If your terms have more than one thing in common you can factor them all out. If you like, for simplicity and until you get used to the idea, you can take them out one by one. No harm there and it will make checking your work easier. With this last one your only flaw (aside from the parenthesis(!)) was that you didn't go far enough. Both terms still had something you could factor out.

    -Dan
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