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Are they raised to fractional powers?
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]
Okay...just give me a sec. I'm only 'half' here. You should notice 2x(3x+4)^(1/3) is a common factor.
yeahhh I can see that. Thanks for trying to help though!
Wait...I made a mistake in the last line...
I'll scan in what I wrote. I never translate things onto the site properly :[
THANK YOU SOOO MUCH! I can see how to do it now, thanks again! :)
wait, isn't 3x+2x = 5x?
It is...damn...thanks for picking that up.
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:\[2x(3x+4)^4/3+x^2*4(3x+4)^1/3\]
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.
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
I am getting ready to post a new thread, maybe you can help?
maybe I can. lets see :)