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Thread: help with some maths

  1. #1

    help with some maths

    I have a brain fade moment with the following equation that I need to factorise by extraction and grouping of a commom factor


    15a - 6b - 5ax + 2bx


    Thanks

  2. #2
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    I think you need the other side of the equation - is this =0, or =y (so then you have four variables).

    Assuming latter (you can always substitute in then:

    a= (y +b(6-2x))/(15-5x)
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  3. #3
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    b=(y+a(5x-15))/(2x-6)
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  4. #4
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    x= (y-15a+6b)/(2b-5a)

    Does this answer your question? If you need to remove the variables further, you can substitute one of the variables with the equation that is it's subject and simplify.....
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  5. #5
    still confused

    one of the others I have is

    5x^4 - 3x^3 + 4x^2 which I have worked out to x^2(5x^2 - 3x +4) with x^2 being the commom factor

    thanks for the quick reply

  6. #6
    15a - 6b - 5ax + 2bx

    to factor
    -(5a-2b)(x-3)

    I think,it's been a while

    if its factor by grouping i would say
    (3-x)(5a-2b)
    Last edited by gelert; 15-01-2015 at 10:52.

  7. #7
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    .....so ..... I *think* I'm down to 2y=x^2-4bx+12b. And now having brain fade myself on the quadratic. Should come out as something like 2y= (x-6b)(x+2b) - but that gives -12b at the end (not plus)

    am I helping, or making things worse?!
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  8. #8
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    Gelerts factorisation works out....

    Your second factorisation also looks fine - you can probably break down the quadratic a little further too - again, I'm struggling with that.
    Last edited by Eric the Red; 15-01-2015 at 10:58.
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  9. #9
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    What are you actually trying to achieve?
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  10. #10
    5x^4 - 3x^3 + 4x^2

    to factorise

    5x4-3x3+4x2

    greatest common factor
    x2(4+-3x+5x2)
    Last edited by gelert; 15-01-2015 at 11:19.

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