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Posted by on Wednesday, December 19, 2012 at 7:15pm.

2x^2-5x-3 show all the step please thanks

  • algebra 2 - , Thursday, December 20, 2012 at 11:39am

    Are you wanting it factorised?

    If so:

    multiply the constant, in this case '-3' by the co-efficient of x^2, so:

    (2x-3=6)

    Take the 2 off the co-efficient and replace the constant, -3, with -6, so you have:

    x^2 -5x -6

    Find products of -6 that also add to make -5.

    The products, in this case, are -6 and 1:

    -6 multiplied by 1 = -6
    -6 + 1 = -5

    You now have both co-efficients of x, so remove the -5 in front of the x in the ORIGINAL equation: '2x^2-5x-3' and replace with '-6x + x' so it becomes:

    2x^2-6x+x-3

    Split the equation between the two co-efficients of x:

    '2x^2-6x' and 'x-3'

    Factorise these separately e.g.

    '2x^2-6x' becomes 2x(x-3).

    and,

    'x-3' becomes 1(x-3).

    As you can see the equation inside both brackets are the same so now take the numbers in front of both brackets:'2x' and '+1' and form a separate bracket: (2x+1)

    Discard one of the (x-3) brackets and keep the other.

    so in brackets: (2x+1) (x-3).

    x=-1/2 or x=3.

    Matt

  • algebra 2 - , Thursday, December 20, 2012 at 11:42am

    In short:
    2x^2-5x-3 becomes 2x^2-6x+x-3.

    Split and factorise for:
    2x(x-3) 1(x-3)

    Which becomes:
    (2x+1)(x-3)

    x=-1/2 or x=3

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