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Math Help - quadratic function general form to standard form

  1. #1
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    quadratic function general form to standard form

    Ok here is what I did. I'm kinda weak from factoring


     f(x)= 2x^2+8x-2

    group:
     f(x)= (2x^2+8x)-2

    Completing the Square:

     f(x)= (2x^2+8x+16-16)-2
     f(x)= (2x^2+8x+16)-16-2
     f(x)= (2x^2+8x+16)-18

    Factoring:

     f(x)= 2(x^2+4x+8)-18

    Where did I go wrong from here?

     f(x)= 2(x+2)-18

    bcoz if I FOIL 2(x+2)(x+2) it will come out:

     2(x^2+4x+4)
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  2. #2
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    Quote Originally Posted by Anemori View Post
    Ok here is what I did. I'm kinda weak from factoring


     f(x)= 2x^2+8x-2

    group:
     f(x)= (2x^2+8x)-2

    Completing the Square:

     f(x)= (2x^2+8x+16-16)-2
     f(x)= (2x^2+8x+16)-16-2
     f(x)= (2x^2+8x+16)-18

    Factoring:

     f(x)= 2(x^2+4x+8)-18

    Where did I go wrong from here?

     f(x)= 2(x+2)-18

    bcoz if I FOIL 2(x+2)(x+2) it will come out:

     2(x^2+4x+4)
    You can't complete the square unless the coefficient of x^2 is 1.


    f(x)= 2x^2+8x-2

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

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

     = 2[(x + 2)^2 - 5]

     = 2(x + 2)^2 - 10.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    You can't complete the square unless the coefficient of x^2 is 1.


    f(x)= 2x^2+8x-2

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

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

     = 2[(x + 2)^2 - 5]

     = 2(x + 2)^2 - 10.


    yea i look back my answer and i made a mistake after the grouping.
    thanks!
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