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Math Help - Factors...and more Factors

  1. #1
    Junior Member fluffy_penguin's Avatar
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    Post Factors...and more Factors

    I hate factors > <

    1. http://i23.tinypic.com/15wflw1.jpg
    Is that correct?

    2. http://i24.tinypic.com/mmx2di.jpg
    Is that correct? Do I have to take it a step further somehow?

    3. http://i24.tinypic.com/106yko7.jpg
    I have no clue what to do here.

    4. http://i23.tinypic.com/2ymi1qa.jpg
    Aside from C and D making no sense. The correct answer would be either A or B but it's B because it's the only one factored right? Isn't A like some GCF factor thing.

    5. http://i20.tinypic.com/29y17jl.jpg
    Is this correct?

    Work:
    (r^2-4tr) + (3wr -12tw)
    gcf:r gcf:3w
    r(r-4t) + 3w(r-4t)
    woot
    =(r-4t)(r+3w)

    6. http://i20.tinypic.com/ztca6g.jpg
    4+3xy-6y-2x

    (4-2x) + (3xy-6y)
    2(2-x) + 3y(x-2)

    so now (2-x) and (x-2) are inverse or whatnot so multiplying one side by -1 would make the one on the right -3y(2-x) or would I leave the 3y alone?

    Edit: got (2-x)(2-3y)
    Last edited by fluffy_penguin; October 5th 2007 at 01:24 PM. Reason: finished problem 6
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, fluffy_penguin!

    Ya done good!


    1) Factor: . 15x^3 + 5x

    My answer: . 5x(3x^2+1) . . . . Right!


    2) Factor: . 7x^6y^4 + 21x^4y^3 + 14xy

    My answer: . 7xy(x^5y^3+3x^3y^2 + 2) . . . . Correct!



    3) Factor out the greatest common factor. Simiplify the factors.
    (4z-3)(z+4) - (4z-3)(z-3)
    Note the common factor: . ({\color{red}4z-3})(z+4) - ({\color{red}4z-3})(z-3)

    Factor it out: . (4z-3)\,\bigg[(z+4) - (z-3)\bigg]

    Simplify: . (4z-3)\,\bigg[z + 4 - z + 3\bigg] \;=\;7(4z-3)



    4) Factor: . rt + rp + nt + np
    Factor by grouping: . r(t+p) + n(t+p)

    Then: . (t+p)(r+n)



    5. Factor:. r^2-1tw + 3wr - 4tr

    Is this correct?
    (r^2-4tr) + (3wr -12tw) \:=\:r(r-4t) + 3w(r-4t) \:=\:(r-4t)(r+3w) . . . . Yes!

    6. Factor: . 4+3xy-6y-2x

    My work:
    (4-2x) + (3xy-6y) \:=\:2(2-x) + 3y(x-2)

    so now (2-x) and (x-2) are inverse or whatnot,
    then I would make the right -3y(2-x) . . . . Good!
    Then you have: . 2(2 - x) - 3y(2 - x)

    Factor out: . (2 - x)(2 - 3y)


    Another approach . . .

    We have: . 4 + 3xy - 6y - 2x

    Rearrange terms: . 4 - 2x - 6y + 3xy

    Factor by grouping: . 2(2 - x) - 3y(2 - x)

    Therefore: . (2 - x)(2 - 3y)

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  3. #3
    Junior Member fluffy_penguin's Avatar
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    Sep 2007
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    Thanks for the help.

    I was going to ask for #3 how you got positive 3 but I seemed to understand it now, the - had to be Multiplied
    to both the z and the -3.

    (4z-3)[(z+4)-(z-3)]
    then
    (4z-3)[z+4-z+3]
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