anonymous
  • anonymous
I am really not sure how to simplify this: 2x(3x+4)^4/3+4x^2(3x+4)^1/3?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Are they raised to fractional powers?
anonymous
  • anonymous
For example,\[2x(3x+4)^{4/3}\]
anonymous
  • anonymous
they are raised to fractional powers....I know there has to be someway the (3x+4) comes out. I'm trying to factor it since there are two of the same terms here. So far this is what I have: [2x(3x+4)^1/3](3x+4)^3/3+2x]

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anonymous
  • anonymous
Okay...just give me a sec. I'm only 'half' here. You should notice 2x(3x+4)^(1/3) is a common factor.
anonymous
  • anonymous
yeahhh I can see that. Thanks for trying to help though!
anonymous
  • anonymous
\[2x(3x+4)^{4/3}+4x^2(3x+2)^{1/3}=2x(3x+2)^{1/3}[(3x+4)+2x]\]\[=2x(3x+2)^{1/3}(6x+4)=4x(3x+2)^{1/3}(3x+2)=4x(3x+2)^{4/3}\]
anonymous
  • anonymous
Wait...I made a mistake in the last line...
anonymous
  • anonymous
I'll scan in what I wrote. I never translate things onto the site properly :[
anonymous
  • anonymous
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anonymous
  • anonymous
THANK YOU SOOO MUCH! I can see how to do it now, thanks again! :)
anonymous
  • anonymous
kk ;]
anonymous
  • anonymous
wait, isn't 3x+2x = 5x?
anonymous
  • anonymous
It is...damn...thanks for picking that up.
anonymous
  • anonymous
heh, np.
anonymous
  • anonymous
Okay...*this* one should be right. Too many distractions going on...
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anonymous
  • anonymous
I just realized something...when I wrote the problem where I have 4x^2, the original problem actually has the x^2 separated from the 4. So it actually is:\[2x(3x+4)^4/3+x^2*4(3x+4)^1/3\]
anonymous
  • anonymous
Does this make a difference to the answer...I went ahead and multiplied x^2+4...to make it 4x^2. I hope this doesn't change all the hard work you did, I just realized this though.
anonymous
  • anonymous
no, it doesn't matter. a * b = b * a. in other words, 7 times 9 is the same as 9 times 7 is the same as 63
anonymous
  • anonymous
okay...thanks.
anonymous
  • anonymous
I am getting ready to post a new thread, maybe you can help?
anonymous
  • anonymous
maybe I can. lets see :)

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