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Math Help - HELP needed with3 algebra problems, factor

  1. #1
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    HELP needed with3 algebra problems, factor

    the section iam doing is called "5-12 solving equations by factoring"
    and i cant seem to get a reasonable answer to these problems

    6c^2-72=11c



    4x^3-12x^2+8x=0




    2n^3-30n^2+100n=0




    please show how you did it so i can learn
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  2. #2
    Junior Member Ruichan's Avatar
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    Quote Originally Posted by elfen_lied View Post
    the section iam doing is called "5-12 solving equations by factoring"
    and i cant seem to get a reasonable answer to these problems

    6c^2-72=11c



    4x^3-12x^2+8x=0




    2n^3-30n^2+100n=0




    please show how you did it so i can learn
    6c^2-72=11c
     6c^2 - 11c - 72 = (2c - 9)(3c + 8);<br />
c = 9/2, <br />
c = -8/3

    4x^3-12x^2+8x=0
     4x(x^2 - 3x + 2) = 4x (x - 1)(x - 2);<br />
x=0, x=1, x=2

    2n^3-30n^2+100n=0
     2n(n^2 - 15n + 50) = 2n(n-5)(n-10);<br />
n=0, n=5, n=10
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by elfen_lied View Post
    6c^2-72=11c
    6c^2-72=11c

    6c^2 - 11c - 72 = 0
    We want a pair of factors of 6 \cdot -72 = -432 (the coefficient of the c^2 term times the coefficient of the constant term) that add to -11:
    1, -432 --> -431
    2, -216 --> -214
    3, -144 --> -141
    etc.

    At some point in the list we get to:
    16, -27 --> -11
    So we split the middle term: -11c = 16c - 27c:
    6c^2 - 11c - 72 = 0

    6c^2  + 16c - 27c - 72 = 0

    2c(3c + 8) - 9(3c + 8) = 0

    (2c - 9)(3c + 8) = 0 (You can verify that this is our factorization by multiplying it out.)

    So either 2c - 9 = 0 or 3c + 8 = 0.

    Thus c = \frac{9}{2} or c = -\frac{8}{3}.

    Quote Originally Posted by elfen_lied View Post
    4x^3-12x^2+8x=0
    4x^3-12x^2+8x=0

    4x(x^2 - 3x + 2) = 0
    You can apply the method above (split the -3x into -x - 2x and factor by grouping as in the above), but you can probably factor this by looking at it. Either way:

    4x(x - 2)(x - 1) = 0

    So x = 0, 1, or 2.


    Quote Originally Posted by elfen_lied View Post
    2n^3-30n^2+100n=0
    2n^3-30n^2+100n=0

    2n(n^2 - 15n + 50) = 0

    To factor the second term we need to find a pair of numbers that are factors of 1 \cdot 50 = 50 that add up to -15. I get -5 and -10:

    2n(n^2 - 15n + 50) = 0

    2n(n^2 - 5n - 10n + 50) = 0

    2n(n(n - 5) - 10(n - 5)) = 0

    2n(n - 10)(n - 5) = 0

    Thus n = 0, 5, or 10.

    -Dan
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    Thanks!
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