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Math Help - General Factoring

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jay123 View Post
    lol...im still lost.

    do i
    X^2-5(2y+3)+6
    X^2-10y-15+6
    X^2-10y+1
    you already had (x - 2)(x - 3), why would you go back to the expanded form? just replace ALL x's in the foiled form with 2y + 3. that's it
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  2. #17
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    sorry. but do replace all (2y+3) with x.?
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  3. #18
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jay123 View Post
    sorry. but do replace all (2y+3) with x.?
    ?? i'm sorry, but i don't see your hang up here


    ({\color {red} x} - 2)({\color {red}x} - 3) = ({\color {red} 2y + 3} - 2)( {\color {red} 2y + 3} - 3)

    now simplify



    This is my 37th post!!!
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  4. #19
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    (x-3)(x-2) (2y+3-2)(2y-3-3) = 2y+1 2y = 2y(2y+1). yessss, thank you..but how do i recognize this as a quad. equation
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  5. #20
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jay123 View Post
    (x-3)(x-2) (2y+3-2)(2y-3-3) = 2y+1 2y = 2y(2y+1). yessss, thank you..but how do i recognize this as a quad. equation
    quadratics are of the form: ax^2 + bx + c

    you had something squared plus a constant times something plus a constant, that is in the form of the quadratic above. in this case you had a = 1, x = 2y + 3, b = -5 and c = 6. so they were in that form


    and it was not an equation, you need an equal sign to be an equation
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  6. #21
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    how about this problem..

    -3x^3-15x^2+27x+135
    i first grouped: (-3x^3-15x^2+27x)+135 = -3x(x^2+5x+9)+135

    so i decided to regroup the 5x+135 = 5(x+27)

    -3x(x^29) = -3x(x+3)(x-3)5(x+27)

    now do i -3x+5 (x+3)(x+27)?
    Last edited by jay123; September 22nd 2007 at 07:50 PM. Reason: exponents
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  7. #22
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jay123 View Post
    how about this problem..

    -3x^3-15x^2+27x+135
    i first grouped: (-3x^3-15x^2+27x)+135 = -3x(x^2+5x+9)+135

    so i decided to regroup the 5x+135 = 5(x+27)

    -3x(x^29) = -3x(x+3)(x-3)5(x+27)

    now do i -3x+5 (x+3)(x+27)?
    this is the same type of question as the first one you asked. there is a common -3x^2 among the first two terms and a common 27 among the last two. factor by grouping
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