(18u^6v^5+24u^3v^3)/42u^2v^5 help??! I don't quite understand how to decide which elements to pull/ factor out etc. step by step explanation would be lovely!
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Originally Posted by lindseyb (18u^6v^5+24u^3v^3)/42u^2v^5 help??! I don't quite understand how to decide which elements to pull/ factor out etc. step by step explanation would be lovely! Numerator: $\displaystyle 18=6\cdot 3$ $\displaystyle 24=6\cdot 4$ $\displaystyle u^6=u^3\cdot u^3$ $\displaystyle v^5=v^3\cdot v^2$ Denominator: $\displaystyle 42=6\cdot 7$ What do you think you should factor out?
The highest common factor of the numerator is $\displaystyle \displaystyle 6u^3v^3$. After you have factorised the numerator, is there anything that cancels with the denominator?
so I pull the 6 out front of them all, then the ^3 in the numerator? which would make it 6(3u^3v^2)^3 6(4uv)3/ 6(7u^2v^5)?
Originally Posted by lindseyb so I pull the 6 out front of them all, then the ^3 in the numerator? which would make it 6(3u^3v^2)^3 6(4uv)3/ 6(7u^2v^5)? $\displaystyle \displaystyle\frac{6u^3v^3(3u^3v^2+4)}{6u^2v^3(7)}$
Alrighty! That makes sooo much more sense! from here can I cancel the 6u^3 to become u? so the final answer would be u(3u^3v^2+4v^2)/ v^3(7)
Originally Posted by lindseyb Alrighty! That makes sooo much more sense! from here can I cancel the 6u^3 to become u? so the final answer would be u(3u^3v^2+4v^2)/ v^3(7) Yes, what about the v?
oh! Can I combine them?? 3v^2+4v^2? so u(3u^7v^2)/ v^3(7)
Originally Posted by lindseyb oh! Can I combine them?? 3v^2+4v^2? so u(3u^7v^2)/ v^3(7) Originally Posted by dwsmith $\displaystyle \displaystyle\frac{uv^3(3u^3v^2+4)}{v^3(7)}$ How about cancel them?
Last edited by dwsmith; Jan 22nd 2011 at 07:26 PM. Reason: forgot u
AH! I'm sorry, I forgot to rewrite the v^3 in the denominator.... that makes sense so the final FINAL result is u(3u^3v^2+4)/ 7
Originally Posted by lindseyb AH! I'm sorry, I forgot to rewrite the v^3 in the denominator.... that makes sense so the final FINAL result is u(3u^3v^2+4)/ 7 Correct.
yay!! Thank you!
one more??! if (x^2-4)/(x^2+4) am I allowed to cancel the x^2's?
Originally Posted by lindseyb one more??! if (x^2-4)/(x^2+4) am I allowed to cancel the x^2's? No. It doesn't simplify.
Alrighty, thank you.
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