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    • Thread Starter
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    Hi guys,

    How would you factorise:

    6x^2+x-2
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    EDIT: Ignore.
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    It's a quadratic so you're trying to write it in the form (ax+b)(cx+d). In particular you can see by multiplying out the factors that you want ac=6, bd=-2 (and ad+bc=1). Try some nice integer values...
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    Take the co-efficient of x^2 and multiply it with the constant, so:

    6 * -2 = -12

    Now find two numbers that multiply to give that -12 and add to give the co-efficient of x (1). Let's say the first number is A and the second number is B. Then write is as

    6x^2 + Ax + Bx - 2 and factorise the 6x^2 + Ax separately from Bx - 2. See how that works out.
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    6x^2+x-2

    6x^2 can be gotten by either using (6x)(x) or (3x)(2x)

    To get -2, the only thing that can be done is (-2)(1) or (2)(-1)

    this leaves us with

    (6x-2)(x+1)=6x^2+4x-2 Incorrect

    (6x+2)(x-1)=6x^2-4x-2 Incorrect

    (3x-2)(2x+1)=6x^2-x-2 Incorrect,as the sign of x in the original equation is negative. Reversing the +ve and-ve signs solves this:



    (3x+2)(2x-1)=6x^2+x-2

    Hope that helped
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    Hmmm,you could do that sum and product thing........

    Its a quadratic so it is in the form ax^2+bx+c

    You times a by c so >6 x -2=-12

    And then basically,you try to find two factors of -12 which will make 1(b).They can be 4 and -3.

    You then split the equation up like this....

    6x^2 +4x-3x-2 and then...you you factorise half of the equation at a time so.....

    2x(3x+2) -1(3x+2)

    The stuff inside the bracket should be the same....so you take one of the brackets..

    (3x+2) and then,the other bracket is just the stuff outside....so (2x-1)

    you are left with(3x+2)(2x-1).


    Hope this helps.
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    ignore
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    which two numbers multiply to make 6x^2

    3x and 2x or 6x and x

    Which two no.s add to make 1 and times to make -2

    2 and -1 or -1 and 2

    (3x + 2) (2x-1)
 
 
 
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