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Math Help - [SOLVED] Help factor 4((m+n)^3)-4 and 36x^2-9y^2+6y-1

  1. #1
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    [SOLVED] Help factor 4((m+n)^3)-4 and 36x^2-9y^2+6y-1

    only 2 i cant get
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  2. #2
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    4(m+n)^3 - 4 =

    4[(m+n)^3 - 1] =

    note the difference of cubes in the brackets ...

    4[(m+n)^3 - 1^3]

    remember how to factor the difference of two cubes?


    36x^2-9y^2+6y-1=36x^2 - (9y^2 - 6y + 1)

    36x^2 - (3y - 1)^2

    difference of two squares, correct?
    Last edited by skeeter; February 12th 2009 at 04:43 PM. Reason: forgot my = sign
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  3. #3
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    Hello, redhawks!


    4(m+n)^3 - 4

    We have: . 4\underbrace{\bigg[(m+n)^3 - 1\bigg] }_{\text{diff. of cubes}}

    . . . = \;\bigg[(m+n)-1\bigg]\,\bigg[(m+n)^2 + (m+n) + 1\bigg]

    . . . = \;(m+n-1)(m^2 + 2mn + n^2 + m + n + 1)



    36x^2 - 9y^2 + 6y - 1

    We have: . 36x^2 - (9y^2 - 6y + 1)

    . . . = \;36x^2 - (9y^2 - 6y + 1)

    . . . = \;\underbrace{36x^2 - (3y - 1)^2}_{\text{diff. of squares}}

    . . . = \;\bigg[6x - (3y-1)\bigg]\,\bigg[6x + (3y - 1)\bigg]

    . . = \;(6x - 3y + 1)(6x + 3y - 1)




    Edit: Too slow . . . again!
    .
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  4. #4
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    Quote Originally Posted by redhawks View Post
    only two i cant get
    4 \left[ (m+n)^3 - 1\right] and factorise the difference of two cubes.

    36x^2 - (9y^2 - 6y + 1) = (6x)^2 - (3y - 1)^2 and factorise the difference of two perfect squares.
    Last edited by mr fantastic; February 12th 2009 at 04:44 PM. Reason: No edit - falgging reply as having been moved from a duplicate posting of the question.
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  5. #5
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    Thanks

    Thanks to both of you
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