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Math Help - Quadratics Factoring 2

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    Quadratics Factoring 2

    Factor: 3(2z + 3)^2 - 9(2z + 3) - 30

    My attempt:

    3(2z + 3)^2 - 9(2z + 3) - 30

    3x^2 - 9x - 30 Replace (2z +3)^2 and (2z +3) with a variable (I choose x)

    3(x^2 - 3x - 10) Find 2 integers whose product is ac = -10 and sum equals b = -3 (+2 and - 5)

    3(x^2 + 2x )(- 5x - 10)

    x(x + 2) -5(x + 2)

    (x - 5) ( x + 2) Plug (2z +3)^2 and (2z +3) back into the x's.

    ((2z + 3)^2) - 5 ((2z + 3) + 2)

    (2z - 2)(2z + 5)

    I got it wrong, the correct answer is 6(z - 1)(2z + 5)

    Where did I go wrong?
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    Re: Quadratics Factoring 2

    We are given to factor:

    3(2z + 3)^2-9(2z + 3)-30

    I would first factor our the 3:

    3((2z + 3)^2-3(2z + 3)-10)

    Now, we need two factors of -10 whose sum is -3, and they are 2 and -5, hence:

    3((2z+3)+2)((2z+3)-5)

    Further simplification will lead to the result you desire...
    Last edited by MarkFL; April 21st 2013 at 12:09 PM.
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    Re: Quadratics Factoring 2

    Why do you need two factors whose sum is -2 and not - 3? The middle term is -3.
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    Re: Quadratics Factoring 2

    Quote Originally Posted by Mathnood768 View Post
    Why do you need two factors whose sum is -2 and not - 3? The middle term is -3.
    Sorry, that was a typo. I have edited my post.
    Last edited by MarkFL; April 21st 2013 at 12:13 PM.
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    Re: Quadratics Factoring 2

    Quote Originally Posted by Mathnood768 View Post
    Factor: 3(2z + 3)^2 - 9(2z + 3) - 30

    My attempt:

    3(2z + 3)^2 - 9(2z + 3) - 30

    3x^2 - 9x - 30 Replace (2z +3)^2 and (2z +3) with a variable (I choose x)

    3(x^2 - 3x - 10) Find 2 integers whose product is ac = -10 and sum equals b = -3 (+2 and - 5)

    3(x^2 + 2x )(- 5x - 10)

    x(x + 2) -5(x + 2)

    (x - 5) ( x + 2) Plug (2z +3)^2 and (2z +3) back into the x's.

    ((2z + 3)^2) - 5 ((2z + 3) + 2)

    (2z - 2)(2z + 5)

    I got it wrong, the correct answer is 6(z - 1)(2z + 5)

    Where did I go wrong?

    3(2z + 3)^2 - 9(2z + 3) - 30

    This is an expression of the form ax2+bx+c so let x = 2z+3 hence 3x2-9x-30

    Multiply 3 by -30 (-90) and find a pair of its factors that sum to -9

    The pair are 6 and -15 so replace -9x with +6x and -15x

    Hence 3x2 +6x -15x -30 now bracket off and factorise

    [(3x2 +6x) -(15x -30)]

    [3x(x+2)-15(x-2)]

    (3x-15)(x+2) = 3(x-5)(x+2)

    Now replace x with 2z+3

    3[(2z+3)-5][(2z+3)+2]

    3(2z-2)(2z+5) = 6(z-1)(2z+5)
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  6. #6
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    Re: Quadratics Factoring 2

    Quote Originally Posted by Mathnood768 View Post
    Factor: 3(2z + 3)^2 - 9(2z + 3) - 30

    My attempt:

    3(2z + 3)^2 - 9(2z + 3) - 30

    3x^2 - 9x - 30 Replace (2z +3)^2 and (2z +3) with a variable (I choose x)

    3(x^2 - 3x - 10) Find 2 integers whose product is ac = -10 and sum equals b = -3 (+2 and - 5)

    3(x^2 + 2x )(- 5x - 10) Should be 3(x^2 + 2x - 5x - 10)

    = 3[ x(x + 2) -5(x + 2) ]

    =3[ (x - 5) ( x + 2) ] Plug (2z +3)^2 and (2z +3) back into the x's.

    No, you just put 2x+3 wherever you see x
    See edits...
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