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Math Help - Having a difficult time with more advanced distributing in algebra?

  1. #1
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    Having a difficult time with more advanced distributing in algebra?

    Hello, I'm in Business with Algebra and Calculus and I know the Calculus pretty well but it's this particular type of distributing that I'm having trouble with. For instance..
    (x^4-2) * 3(x^2+2)^2 *2x + (x^2+2)^3 * 4x^3

    There is just so many numbers I can't figure out which numbers and values I should take out

    The answer to it is:
    2x(x^2+2)^2[3x^4 - 6 + 2x^4 + 4x^2]

    Then:
    2x(x^2 +2)^2 * (5x^4 + 4x^2 - 6)


    I just need to figure out how to do this then I think I will be good for my exam.
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  2. #2
    Super Member Quacky's Avatar
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    Re: Having a difficult time with more advanced distributing in algebra?

    \underbrace{(x^4-2)\times{3(x^2+2)^2}\times{2x}} + (x^2+2)^3\times{4x^3}
    In the context I will be using, this is going to be seen as one term, as everything is connected through multiplication. Everything after the + that follows constitutes the second term.


    Look for common factors to both terms of the expression. I've highlighted some below:

    (x^4-2)\times{3(x^2+2)^2}\times{{\color{red}2x}} + (x^2+2)^3\times{{\color{red}4x^3}}

    Both terms in red are multiples of 2x - the latter is 2\times{2}\times{x}\times{x}\times{x}

    So we can take this out as a factor:

    =2x[

    And then we just need to divide each term by 2x once - almost everything stays the same, and we have:

    =2x[(x^4-2)\times{3(x^2+2)^2} + (x^2+2)^3\times{2x^2}] as the only terms that have changed are the terms in red.

    We can then look for another common factor - and indeed there is one.

    =2x[(x^4-2)\times{3{\color{red}(x^2+2)^2}} + {\color{red}(x^2+2)^3}\times{2x^2}]

    We can take out a factor of (x^2+2)^2 as this is the highest power of (x^2+2) that is common to both terms.

    =2x(x^2+2)^2[

    And again, we divide through each term by the common factor once.

    =2x(x^2+2)^2~[(x^4-2)\times{3} + (x^2+2)\times{2x^2}]

    Notice that the last term still has (x^2+2)^1 remaining, because it was initially to the power 3 and only the second power was common to both terms.

    Now it's a case of simplifying what's left in the square brackets - hopefully you'll be okay with that.
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  3. #3
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    Re: Having a difficult time with more advanced distributing in algebra?

    Thanks for the very informative answer, I think I'm beginning to understand it now. The plus sign separating the two terms explained a lot. I think I just have to practice these sort of problems some more now . BTW, how did you show your answer in handwriting-like form?
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  4. #4
    Super Member Quacky's Avatar
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    Re: Having a difficult time with more advanced distributing in algebra?

    I hope it helped. To type the mathematical notation requires an understanding of latex, but basically, using the [Tex][\tex] tags at the top right of the page, you can do the following:

    [TEX]x^2+7x+9[\TEX] gives x^2+7x+9

    The codes for other symbols vary, but you'll get used to them in time.
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  5. #5
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    Re: Having a difficult time with more advanced distributing in algebra?

    Ah I see, very cool. I'll use this in the future.
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