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    (Original post by eliotball)
    No problem, I'm sure you had good intentions posting that solution, it was just a pretty awful mistake :P
    Um yeah, big mistake! I feel stupid now, haha.
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    (Original post by kingkongjaffa)
    thanks how do you come to that? I'm not very good at factorising with things that have coefficients of x :P
    You have to think that you have

    3x^2 -10x +7 = 0

    So something x times x gives 3x^2, so that something can only be 3. That gives you a starting point with two numbers that are unknown (we'll call them a and b)
    (3x + a)(x + b)= 0
    You also know that if you seperate out this bracket (which I am sure you can do)
    You get...
    3x^2 + ax + 3bx + ab = 0

    So a times be has to be 7. Leaves you with a choice of 1, -1, 7 and -7 for a and b.
    You also know that ax + 3bx is -10x, so then you know that a + b is -10. There is only one solution of the a and bs that we have shown will work.

    Hope that helps!
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    (Original post by birduk)
    You have to think that you have

    3x^2 -10x +7 = 0

    So something x times x gives 3x^2, so that something can only be 3. That gives you a starting point with two numbers that are unknown (we'll call them a and b)
    (3x + a)(x + b)= 0
    You also know that if you seperate out this bracket (which I am sure you can do)
    You get...
    3x^2 + ax + 3bx + ab = 0

    So a times be has to be 7. Leaves you with a choice of 1, -1, 7 and -7 for a and b.
    You also know that ax + 3bx is -10x, so then you know that a + b is -10. There is only one solution of the a and bs that we have shown will work.

    Hope that helps!
    thanks a lot
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    (Original post by kingkongjaffa)
    ooh can you explain this please firstly AC is 3*7 but can you explain why you did that.
    secondly can you explain how you go from 3x(x-1) -7(x-1) to (3x-7) (x-1)?
    Well because the co-efficient of x^2 is greater than two, you must multiply the co-efficient of x^2 by the constant which is A x C ( if confused as to the letters, remember the quadratic equation).

    You go from 3x(x-1) -7(x-1) to (3x-7) (x-1) by eliminating one (x-1) and making the numbers outside each bracket into another bracket. REMEMBER to eliminate one bracket, both brackets have to be the same.....

    eg: 4x(x+5) 5(x+5) will become (4x +5)(x+5)

    and, 11x(x-3) -4(x-3) = (11x-4)(x-3)


    Note the Solutions for x exist when you make the bracket equal to 0...

    eg (4x + 5)(x+5) =0

    4x+5=0
    4x= -5
    x= -5/4

    and, x +5 = 0
    x = -5
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    (Original post by Tizmo)
    Um yeah, big mistake! I feel stupid now, haha.
    The feeling will wear off :cool:
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    (Original post by crazycake93)
    Well because the co-efficient of x^2 is greater than two, you must multiply the co-efficient of x^2 by the constant which is A x C ( if confused as to the letters, remember the quadratic equation).

    You go from 3x(x-1) -7(x-1) to (3x-7) (x-1) by eliminating one (x-1) and making the numbers outside each bracket into another bracket. REMEMBER to eliminate one bracket, both brackets have to be the same.....

    eg: 4x(x+5) 5(x+5) will become (4x +5)(x+5)

    and, 11x(x-3) -4(x-3) = (11x-4)(x-3)


    Note the Solutions for x exist when you make the bracket equal to 0...

    eg (4x + 5)(x+5) =0

    4x+5=0
    4x= -5
    x= -5/4

    and, x +5 = 0
    x = -5

    Thanks i understand it now the examples helped aswell thanks for taking the time to read the thread, all of you.
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