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Math Help - factoring (common factoring complex trinomials)

  1. #1
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    factoring (common factoring complex trinomials)

    so the exuation is :

    2x^2+18x+40


    I know I have to common factor first, but I don't know exaclly where to go from there. It's grade 10 accademic math..I duuno if that means anything?

    so I put it so it was like this:


    2(x^2+9x+20)



    and I duuno where to go from there?

    Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by smallman_11 View Post
    so the exuation is :

    2x^2+18x+40


    I know I have to common factor first, but I don't know exaclly where to go from there. It's grade 10 accademic math..I duuno if that means anything?

    so I put it so it was like this:


    2(x^2+9x+20)



    and I duuno where to go from there?

    Any help would be appreciated.
    And how does that factor?
    x^2+9x+20=(x+4)(x+5)
    Thus the answer is,
    2(x+4)(x+5).
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    Thank you
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  4. #4
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    Quote Originally Posted by smallman_11 View Post
    ...
    2(x^2+9x+20)

    and I duuno where to go from there?
    Any help would be appreciated.
    Hello, smallman,

    I presume that you are a little bit uncertain how to factorize
    x^2+9x+20

    To do this probelm you must know the binomial formulae:
    (a+b) = a + 2ab + b [1]
    (a-b) = a - 2ab + b [2]
    (a+b)(a-b) = a - b [3]

    The formula [3] is very important with your problem:
    1. Complete square with the first two summands by adding a third summand and subtract this summand immediately:
    x^2+9x+20 = x^2 + 9x +\left( \frac{9}{2} \right)^2-\left( \frac{9}{2} \right)^2 +20 = \left( x^2 + 9x +\left( \frac{9}{2} \right)^2 \right)-\left( \frac{9}{2} \right)^2 +20
    2. Unite the first three summands to a binom [1] and the last two summands to a square:
    \left( x +\left( \frac{9}{2} \right) \right)^2-\left( \frac{1}{4} \right)

    3. Now you've got a term similar to the RHS of formula [3]. Use the LHS of this formula:

    \left( x +\left( \frac{9}{2} \right) \right)^2-\left( \frac{1}{2} \right)^2 = \left(x +\left( \frac{9}{2} \right)+\left( \frac{1}{2} \right)  \right) \cdot \left(x +\left( \frac{9}{2} \right)-\left( \frac{1}{2} \right)  \right)

    4. Finally you get:

    x^2 + 9x +20 = \left(x +\left( \frac{9}{2} \right)+\left( \frac{1}{2} \right)  \right) \cdot \left(x +\left( \frac{9}{2} \right)-\left( \frac{1}{2} \right)  \right) = (x+5)(x+4)

    That's the TPH's result.

    EB
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