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Math Help - Factoring complex trinomials

  1. #1
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    Factoring complex trinomials

    6x^2+6^x-12

    I get an answer that works of (6x-6)(x+2) but the book says
    (2x+4)(3x-3). What do I have to do to match the book or what did I do wrong?
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  2. #2
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    Re: Factoring complex trinomials

    Quote Originally Posted by Capnzims View Post
    6x^2+6^x-12

    I get an answer that works of (6x-6)(x+2) but the book says
    (2x+4)(3x-3). What do I have to do to match the book or what did I do wrong?
    6x^2+6x-12=6(x^2+x-2)=2\cdot 3 \cdot(x+2)(x-1)=(2x+4)(3x-3)
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  3. #3
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    Re: Factoring complex trinomials

    So factor the 6 outside?
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    Re: Factoring complex trinomials

    yes
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  5. #5
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    Re: Factoring complex trinomials

    Apply the 2 to the first () and the 3 to the next ()? Can I always factor like that first before working the problem on trinomials?
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    Re: Factoring complex trinomials

    yes
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  7. #7
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    Re: Factoring complex trinomials

    Quote Originally Posted by Capnzims View Post
    6x^2+6^x-12

    I get an answer that works of (6x-6)(x+2) but the book says
    (2x+4)(3x-3). What do I have to do to match the book or what did I do wrong?
    Hi Capnzims,

    When asked to factor a polynomial that has a common factor over all terms, you should always factor that out first. Then, factor the remaining polynomial, if possible.

    \displaystyle {6x^2+6x-12=6(x^2+x-2)=6(x+2)(x-1)}

    You wouldn't use \displaystyle {(2x+4)(3x-3)} because there are common factors in each binomial.
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  8. #8
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    Re: Factoring complex trinomials

    Now, do you understand that 6(x- 1)(x+ 2)= (6x- 6)(x+2)= (2x+ 4)(3x- 3)?

    If you multiply the "6" into x- 1, you get (6x- 6)(x+ 2). If, instead, you factor 6 into (2)(3), multiply the "3" into (x- 1) and the "2" into x+ 2, you get (2x+ 4)(3x- 3).
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