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    how do you factorise : 6a^2 x-abx-12b^2 x
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    I'm not a 100% sure but this is what I would do:

    You have: y = 6a^2x - abx - 12b^2x
    First thing I would do is take x out:
    y = x (6a^2 - ab - 12b^2)

    Then you can take two approaches, either factorize it regularily, by taking (a - b) (a - b) and respectfully changing the signs and the number or you could use the formula:

    (-b (plus/minus)(square root of b^2 - 4ac)) / 2a = x

    In this case you have the variables: a = 6, b = -1 and c = -12, as seen in your equation above.

    If you plug in the values you will get:

    x = 1 (plus/minus) (root of: (-1)^2 - 4*6*(-12)) / 2*6
    x = 1 (plus/minus) (root of: 1 + 288) / 12
    x = 1 (plus/minus) (root of: 289) / 12

    you can find the root of 289, which is 17 and put it in:

    x = 1 (plus/minus) 17 / 12

    From this you will get two values:

    x = (1 + 17)/ 12 and x = (1 - 17)/ 12

    Thus you will get two values for x, being x = 3/2 and x = -4/3

    Thus;

    y = x (a -3/2b) (a + 4/3b)

    or, more neatly:

    y = x (2a -3b) (3a + 4b)

    If you want to check it would give:

    y = x ((2a)(3a)+(2a)(4b)+(-3b)(3a)+(-3b)(4b))
    y = x (6a^2 + 8ab - 9ab -12b^2)
    y = x (6a^2 - ab -12b^2)

    And

    y = 6a^2x - abx -12b^2x

    I hope this helped.
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    (Original post by emmalav)
    how do you factorise : 6a^2 x-abx-12b^2 x
    Take the x out and you are left with

    6a^2 - ab - 12b^2

    Are you familiar with the method 6 times -12 = -72

    Then -9 x 8 = -72
    And -9 x 8 = -1

    So you have

    x(6a^2 - 9ab + 8ab - 12b^2) = x(3a(2a -3b) + 4b(2a - 3b)) = x(3a+4b)(2a-3b)
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    ^^ cant just take out x, the x is still there, need to write it in
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    (Original post by TenOfThem)
    Of course it is still there

    I do not believe that I suggested otherwise

    I showed a method not an answer
    but your last line read (3a+4b)(2a-3b) ... where was the x?
    obviously i can see that you knew it but dont leave it out yeh?
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    (Original post by Blake-inator)
    ...
    (Original post by Mr M)
    ...
    Happy now?

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    (Original post by TenOfThem)
    Happy now?

    For now but make sure you don't leave it out next time yeh?
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    (Original post by Mr M)
    For now but make sure you don't leave it out next time yeh?
    yeh

    I is lurnt me lesson innit
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    (Original post by TenOfThem)
    yeh

    I is lurnt me lesson innit
    leave it out guv!
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    (Original post by Dornishman)
    I'm not a 100% sure but this is what I would do:

    You have: y = 6a^2x - abx - 12b^2x
    First thing I would do is take x out:
    y = x (6a^2 - ab - 12b^2)

    Then you can take two approaches, either factorize it regularily, by taking (a - b) (a - b) and respectfully changing the signs and the number or you could use the formula:

    (-b (plus/minus)(square root of b^2 - 4ac)) / 2a = x

    In this case you have the variables: a = 6, b = -1 and c = -12, as seen in your equation above.

    If you plug in the values you will get:

    x = 1 (plus/minus) (root of: (-1)^2 - 4*6*(-12)) / 2*6
    x = 1 (plus/minus) (root of: 1 + 288) / 12
    x = 1 (plus/minus) (root of: 289) / 12

    you can find the root of 289, which is 17 and put it in:

    x = 1 (plus/minus) 17 / 12

    From this you will get two values:

    x = (1 + 17)/ 12 and x = (1 - 17)/ 12

    Thus you will get two values for x, being x = 3/2 and x = -4/3

    Thus;

    y = x (a -3/2b) (a + 4/3b)

    or, more neatly:

    y = x (2a -3b) (3a + 4b)

    If you want to check it would give:

    y = x ((2a)(3a)+(2a)(4b)+(-3b)(3a)+(-3b)(4b))
    y = x (6a^2 + 8ab - 9ab -12b^2)
    y = x (6a^2 - ab -12b^2)

    And

    y = 6a^2x - abx -12b^2x

    I hope this helped.
    thank you
 
 
 
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