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Thread: Factoring Trinomials...I think?

  1. #1
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    Unhappy Factoring Trinomials...I think?

    Hello all. I've been working on this stuff all day...and I just thought to Google math forums, and ended up here! Well..there are a couple of problems that differ slightly from each other that I'm having problems with. The first one is...

    12x+34x+24

    I believe the first step is to find the GCF, which is 6, and divide.

    6x (2x+17+4)

    -sigh- This is as far as I get though...a little help would be so very much appreciated.
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  2. #2
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    Quote Originally Posted by burgessjkkb View Post

    I believe the first step is to find the GCF, which is 6, and divide.

    6x (2x+17+4)
    Correct, although you now have $\displaystyle 6 (2x^2+17x+4)$ not $\displaystyle 6x (2x+17+4)$ . The $\displaystyle x$ term is not common to take out.

    To factor $\displaystyle 2x^2+17x+4$ look for a solution (trial and error) such that $\displaystyle 2x^2+17x+4 = (2x+a)(x+b)$ where a and b multiply to give 4.

    This must also expand to give $\displaystyle 17x$ as the middle term. Could take some time. Good luck!
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  3. #3
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    First of all, $\displaystyle 6 \cdot 17 \neq 34$.

    In fact, $\displaystyle 6$ does not go into $\displaystyle 34$.

    The highest common factor is actually $\displaystyle 2$.


    So $\displaystyle 12x^2 + 34x + 24 = 2(6x^2 + 17x + 12)$.


    Now, instead of using trial and error, find two numbers that multiply to become $\displaystyle 6 \cdot 12 = 72$ and add to become $\displaystyle 17$.

    I think you'll find that the numbers are $\displaystyle 9$ and $\displaystyle 8$.


    So $\displaystyle 2(6x^2 + 17x + 12) = 2(6x^2 + 9x + 8x + 12)$

    $\displaystyle = 2[3x(2x + 3) + 4(2x + 3)]$

    $\displaystyle = 2(2x + 3)(3x + 4)$.
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