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    I have to use the completed square form to factorise 6x^2 - 5x -6. How would I do it?

    I tried factorising 6x^2 - 5x to 6 (x^2 -5/6 x) and then completing the square for that. I'm not 100% sure if there's a better method because I end up with lots of fractions and roots when the answer is apparently (3x + 2)(2x - 3).
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    (Original post by Coco_Monkey)
    I have to use the completed square form to factorise 6x^2 - 5x -6. How would I do it?

    I tried factorising 6x^2 - 5x to 6 (x^2 -5/6 x) and then completing the square for that. I'm not 100% sure if there's a better method because I end up with lots of fractions and roots when the answer is apparently (3x + 2)(2x - 3).
    If the answer is (3x+2)(2x-3) then you shouldn't be trying to complete the square. That's just factorising into two brackets
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    (Original post by AHappyStudent)
    If the answer is (3x+2)(2x-3) then you shouldn't be trying to complete the square. That's just factorising into two brackets
    I think the question is asking me to complete the square and to then factorise from that answer, so having to go by a specific method
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    If you 'have' to use completing the square, then your method is the most straightforward method to do so. If you are allowed to factorise it with brackets, then doing so would be a lot easier.
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    (Original post by Coco_Monkey)
    I think the question is asking me to complete the square and to then factorise from that answer, so having to go by a specific method
    Can you post your working?
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    (Original post by Coco_Monkey)
    I think the question is asking me to complete the square and to then factorise from that answer, so having to go by a specific method
    you would put 6 outside the bracket and the rest of the equation inside and then complete the square.
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    6x^2 - 5x -6
    6(x^2-5x-6)
    6((x-5/2)^2-5/2^2-6)
    carry on from here
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    (Original post by Sonechka)
    Can you post your working?
    Here:
    Attached Images
     
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    (Original post by madi24)
    6x^2 - 5x -6
    6(x^2-5x-6)
    6((x-5/2)^2-5/2^2-6)
    carry on from here
    I don't understand?
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    (Original post by Coco_Monkey)
    Here:
    \displaystyle 6\left(x-\frac{5}{12}\right)^2-\frac{1014}{144}

    From here instead do this

    \displaystyle =6\left(\frac{12x-5}{12}\right)^2-\frac{1014}{144}

    \displaystyle =\frac{ 6\left( 12x-5\right)^2 }{144}-\frac{1014}{144}

    Now take out a factor of \frac{6}{144} and then use difference of two squares.

    But it may even be easier to just solve the quadratic using the completing the square form and then work backwards to factorise it.

    Are you 100% sure that you were asked to factorise this using completing the square? It is extremely inefficient.
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    (Original post by Notnek)
    \displaystyle 6\left(x-\frac{5}{12}\right)^2-\frac{1014}{144}

    From here instead do this

    \displaystyle =6\left(\frac{12x-5}{12}\right)^2-\frac{1014}{144}

    \displaystyle =\frac{ 6\left( 12x-5\right)^2 }{144}-\frac{1014}{144}

    Now take out a factor of \frac{6}{144} and then use difference of two squares.

    But it may even be easier to just solve the quadratic using the completing the square form and then work backwards to factorise it.

    Are you 100% sure that you were asked to factorise this using completing the square? It is extremely inefficient.
    I've done what you said, but I don't know if I've gone wrong or where to go next. I got to 6/144 (12x + 8)(12x -18)
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    (Original post by Coco_Monkey)
    I've done what you said, but I don't know if I've gone wrong or where to go next. I got to 6/144 (12x + 8)(12x -18)
    You're close. Next you can take out a factor of 4 from the (12x + 8) bracket and a factor of 6 from the (12x - 18) bracket.
 
 
 
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