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Math Help - General Math Questions, exponents

  1. #1
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    General Math Questions, exponents

    Hey, I'm working on my math homework and I've come across several questions that I was hoping someone could please clarify for me.

    1. After simplifying the addition I came up with an answer that has x^7 + x^6 + x^5 etc....(with coefficients in front as well, just simplified it for post) all divided by (1- 2x^4) --> is this fully simplified? I would like to factor it further but the only way I know how to simplify it further is through the long trial and error method of finding one factor than using long division.

    2. I have a question that has cuberoot (x + 1) * 3(cuberoot (x+1)^2)
    Am I allowed to just multiply it through so that I get (x+1)^3 and then cancel the three with the cuberoot or is there some other method that I am supposed to use.

    3. (similar to my first question) --> the question asks to simplify and after doing a couple things I'm stuck at:

    (10x^7 - 5 + 15x^3 + 20x^5)/ x^4

    I don't think this is fully simplified but I'm unsure of how to combine or factor these piece parts.

    The original question was:
    10x^3 - 5x^(-4) + 15x^(-1) + 20x

    4. Am I allowed to split up division: for example: (2/5 + 2/4)/5 = (2/5)/5 + (2/4)/5 ??? (In the cases i tried it worked but I figured I would ask the question anyways)

    Sorry for long post (I keep adding as I go) Please help!!!!!


    thanks, bye
    Last edited by Mudd_101; January 19th 2009 at 11:27 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mudd_101 View Post
    Hey, I'm working on my math homework and I've come across several questions that I was hoping someone could please clarify for me.

    1. After simplifying the addition I came up with an answer that has x^7 + x^6 + x^5 etc....(with coefficients in front as well, just simplified it for post) all divided by (1- 2x^4) --> is this fully simplified? I would like to factor it further but the only way I know how to simplify it further is through the long trial and error method of finding one factor than using long division.
    you can factor x^5 from the top. and the bottom is the difference of two squares. you can actually apply the difference of two squares formula twice. but i don't think they'd want you to mess with the denominator, since it doesn't look simpler and will end up involving radicals.

    wait, when you say "etc" what do you mean? please type out the whole question. does it end with 1? something else?
    2. I have a question that has cuberoot (x + 1) * 3(cuberoot (x+1)^2)
    Am I allowed to just multiply it through so that I get (x+1)^3 and then cancel the three with the cuberoot or is there some other method that I am supposed to use.
    you asked this in someone else's thread and it was answered. makes me wonder if that is the case with the rest of these problems.

    3. (similar to my first question) --> the question asks to simplify and after doing a couple things I'm stuck at:

    (10x^7 - 5 + 15x^3 + 20x^5)/ x^4

    I don't think this is fully simplified but I'm unsure of how to combine or factor these piece parts.

    The original question was:
    10x^3 - 5x^(-4) + 15x^(-1) + 20x
    just factor 5x^{-4} from the expression.


    4. Am I allowed to split up division: for example: (2/5 + 2/4)/5 = (2/5)/5 + (2/4)/5 ??? (In the cases i tried it worked but I figured I would ask the question anyways)

    Sorry for long post (I keep adding as I go) Please help!!!!!


    thanks, bye
    yes, doing that is valid. what did you end up with? i would have probably combined the fractions in the top first, but i guess it doesn't matter. you would be expected to end up with one fraction in the end anyway
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  3. #3
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    Okay uhhh.............

    for the first question the question is:
    (28x^7 -22x^6 + 42x^5 - 42x^4 - 24x^3 - 18x^2 + 2x + 2)/(1-2x)^4

    Yeah I guess my bad what I meant by the etc..... was from x^7 down to x^1

    ---> the expression in the first question is what I have so far after already simplifying a more complex equation and I don't know if I'm done or not.



    2. !!!!!!!!!!!!!! How did you know that????? I thought it would be more convenient to put all my questions in one post so I reposted it I will check the other thread. --> that's the only question I reposted


    I checked the other thread They Took down my POST!@@!@@#!@@#@ Can you answer it please it turns out they gave me an Infraction!!!!!!!!!!!!!!!!!!!!!!!!!!!

    3. so fully simplified form is: (5x^(-4))(2x^7 - 1 + 3x^3 + 4x^5)
    Is it not possible to do anything with the exponents inside the brackets?




    Thanks
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mudd_101 View Post
    Okay uhhh.............

    for the first question the question is:
    (28x^7 -22x^6 + 42x^5 - 42x^4 - 24x^3 - 18x^2 + 2x + 2)/(1-2x)^4

    Yeah I guess my bad what I meant by the etc..... was from x^7 down to x^1

    ---> the expression in the first question is what I have so far after already simplifying a more complex equation and I don't know if I'm done or not.
    this is a weird question. i don't see much else to do besides factoring out a 2 from the numerator. i doubt it can go simpler, but maybe i'm too tired to see.

    2. !!!!!!!!!!!!!! How did you know that?????
    i was the one that answered it in the other thread

    I thought it would be more convenient to put all my questions in one post so I reposted it I will check the other thread. --> that's the only question I reposted
    never repost. even if you're grouping it with other stuff. it is the same as double posting.

    I checked the other thread They Took down my POST!@@!@@#!@@#@ Can you answer it please it turns out they gave me an Infraction!!!!!!!!!!!!!!!!!!!!!!!!!!!
    *sigh*

    i will give you the idea behind it

    lets say you have \sqrt[3]{x} \cdot \sqrt[3]{x^2}, then that is the same as

    \sqrt[3]{x} \cdot ( \sqrt[3]{x})^2 = ( \sqrt[3]{x} )^3 = x

    3. so fully simplified form is: (5x^(-4))(2x^7 - 1 + 3x^3 + 4x^5)
    Is it not possible to do anything with the exponents inside the brackets?
    Thanks
    nope, doesn't look that way. that's it
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  5. #5
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    Thanks Jhevon !!!! Greatly Appreciated
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mudd_101 View Post
    Thanks Jhevon !!!! Greatly Appreciated
    you're welcome
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