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Math Help - I'm totally confused!

  1. #1
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    Unhappy I'm totally confused!

    I am so confused on these!!

    I get stated and then don't know where to start and where to end. Please help.

    6a-3a^2-2ab+a^2b

    I started with 3(3a-a^2) but not even sure if that is correct.

    Thank you
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  2. #2
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    Hi

    6a-3a^2-2ab+a^2b = 2a(3-b)+a^2(b-3) = (a^2-2a)(b-3) = a(a-2)(b-3)
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by tpoma View Post
    I am so confused on these!!

    I get stated and then don't know where to start and where to end. Please help.

    6a-3a^2-2ab+a^2b

    I started with 3(3a-a^2) but not even sure if that is correct.

    Thank you
    First, factor out an a, then try gouping the terms when you have a 4-term polynomial.

    Group the first 2, then the second 2 and pull out common factors.

    6a-3a^2-2ab+a^2b

    a(6-3a-2b+ab)

    a [ (6-3a)-(2b-ab) ]

    a [ 3(2-a)-b(2-a) ]

    How's this working out for you? Can you finish up?

    I'll finish up since running-gag had a slightly different approach

    a(2-a)(3-b) \ \ or \ \ a(a-2)(b-3)
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  4. #4
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    Red face

    Ok, I am trying this again, can you please tell me if this is correct or anywhere close to what I am trying to accomplish? Thanks

    4k-8-k^3+2k^2

    I have come up with 2k(2+k^2+2k) I don't feel confident that it is anywhere close to what I am suppose to have.
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  5. #5
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    No

    4k-8-k^3+2k^2 = 4(k-2) - k^2(k-2) = (4 - k^2)(k-2)

    4k-8-k^3+2k^2 = (2+k)(2-k)(k-2) = -(k+2)(k-2)^2
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  6. #6
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    Exclamation

    Quick question,

    On this problem why is it that the 9 seems to disapear? It seems to just drop off????


    2y^3-18y^2= -28y

    My final answer was 2y(y-2)(y-7)=0
    2y=0 y-2=0
    y=0 y=2

    y-7=0 y=7
    Where did the 9 go to?? I don't understand it but I think I got it right.

    Thanks
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  7. #7
    A riddle wrapped in an enigma
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    Quote Originally Posted by tpoma View Post
    Quick question,

    On this problem why is it that the 9 seems to disapear? It seems to just drop off????


    2y^3-18y^2= -28y

    My final answer was 2y(y-2)(y-7)=0
    2y=0 y-2=0
    y=0 y=2

    y-7=0 y=7
    Where did the 9 go to?? I don't understand it but I think I got it right.

    Thanks
    Your answers are correct. y = {0, 2, 7}

    Here's how it goes:

    2y^3-18y^2=-28y

    2y^3-18y^2+28y=0

    2y(y^2-9y+14)=0

    When you factor the trinomial, the sum of the two factors that make 14 must add to make -9.

    2y(y-7)(y-2)=0

    You see. The -9y is made up of -7y + -2y. You may see it better this way.

    2y(y^2-7y-2y+14)=0

    Here, I just replaced the -9y in the middle with the sum of the two factors that made +14. Now, if I group the 4 term polynomial, I arrive at the desired factored result.

    2y[(y^2-7y)-(2y-14)]

    2y[y(y-7)-2(y-7)]

    2y(y-7)(y-2)=0

    2y=0 \ \ or \ \ y-7=0 \ \ or \ \ y-2=0

    y=0 \ \ or \ \ y=7 \ \ or \ \ y=2
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  8. #8
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    Talking

    ahhhh got ya, so I could have used FOIL on this problem once I got the equation set up....didn't even see it in a FOIL format! Thank you for your explination! They always seem to help when I am having problems!!
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