- anonymous

Simplify:

- schrodinger

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- anonymous

|dw:1440898252857:dw|

- LynFran

not clear...

- anonymous

\[(4-x^2)(1/3)(6x+1)^{-2/3}(6)-(6x+1)^{1/3}(-2x) / (4-x^2)^2\]

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## More answers

- LynFran

6*1/3=2 ...\[\frac{[2( 4-x ^{2})(6x+1)^{\frac{ -2 }{ 3 }}]-[(6x+1)^{\frac{ 1 }{ 3 }}(2x)] }{ (4-x ^{2})^{2} }\]

- anonymous

Do you start canceling?

- anonymous

I have the answer, and that isn't right. I need to know how to work it.

- LynFran

ok so we have ...

- LynFran

\[\frac{ 2(6x+1)^{-1/3} [(6x+1)^{-1/3}(4-x ^{2})-(x)]}{ (4-x ^{2})^{2} }\]

- LynFran

@Loser66 what do u think..

- Nnesha

1/3 is an exponent at the numerator ?

- anonymous

yes

- Nnesha

\[\large\rm \frac{ 6 (4-x^2)^\frac{ 1 }{ 3 } (6x+1)^\frac{ -2 }{ 3}-(6x+1)^\frac{ 1 }{ 3 }(-2x) }{ (4-x^2)^2}\]hmm like this ?

- anonymous

There is no 1/3 exponent for (4-x^2).

- anonymous

\[(4-x^2)(1/3)(6x+1)^{-2/3}-(6x+1)^{1/3}(-2x) / (4-x^2)^2\] is the problem

- Nnesha

ohh i asked before

- Nnesha

\[\large\rm \frac{ 6(\frac{1}{3}) (4-x^2) (6x+1)^\frac{ -2 }{ 3}-(6x+1)^\frac{ 1 }{ 3 }(-2x) }{ (4-x^2)^2}\]
looks right ?

- anonymous

No

- Nnesha

no hmm can you please take a screenshot

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @ducksonquack
\[(4-x^2)(1/3)(6x+1)^{-2/3}(6)-(6x+1)^{1/3}(-2x) / (4-x^2)^2\]
\(\color{blue}{\text{End of Quote}}\)
there is 6

- anonymous

##### 1 Attachment

- Nnesha

alright thanks

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @ducksonquack
I have the answer, and that isn't right. I need to know how to work it.
\(\color{blue}{\text{End of Quote}}\)
what's the answer
jsut want to check if i'm doing this right or not>.<

- anonymous

\[2(5x^2+x+4) / (4-x^2)^2(6x+?)^{2/3}\]

- anonymous

Didn't catch what the ? mark was when we were given the answers.

- Jhannybean

Well that's in your problem, it states \((6x+1)\) in parenthesis.... so im sure the \(? = 1\)

- anonymous

Okay. What's next?

- Jhannybean

One second im solving it

- Jhannybean

Alright got it, so...

- anonymous

?

- Jhannybean

\[=\frac{\frac{1}{3}\cdot 6 \cdot (4-x^2)(6x+1)^{-2/3}-(6x+1)^{1/3}(-2x)}{(4-x^2)^2}\]\[=\frac{(6x+1)^{-2/3}\left(2\left[4-x^2-(6x+1)(-x)\right]\right)}{(4-x^2)^2}\]\[=\frac{2\left[4-x^2+6x+x\right]}{(4-x^2)^2(6x+1)^{2/3}}\]\[=\boxed{\frac{2(5x^2+x+4)}{(4-x^2)^2(6x+1)^{2/3}}}\]

- anonymous

Can you explain it to me step by step?

- Jhannybean

Okay so what you want to do first is look for a common like term. in this case it would be \((6x+1)^{-2/3}\). Notice how its a common multiplier in the numerator? We can pull it out. Now I pulled out \((6x+1)^{-2/3}\) instead of the \((6x+1)^{1/3}\) because outof the two, \((6x+1)^{-2/3}\) is the smaller power, therefore pulling it out as a common factor, I am left with \((6x+1)^1\) inside the brackets. (Remember: \((6x+1)^{-2/3} \cdot (6x+1)^1 = (6x+1)^{1/3}\))
Now I pulled out a 2 inside the brackets because that is another common like term between the entire function within the brackets. \([2(4-x^2)-(6x+1)(-2x)] \iff 2[(4-x^2-(6x+1)(-x)]\)
Then I simplify my fraction even further by turning \((6x+1)^{-2/3}\) positive by putting it in the denominator. this turns it positive. Recall that: \(x^{-\#} \iff \dfrac{1}{x^{\#}}\)
So far we have: \[\frac{2[4-x^2-(6x+1)(-x)]}{(4-x^2)^2(6x+1)^{2/3}}\] Now we simplify the stuff inside the brackets without multiplying in the 2. \[\frac{2[4-x^2 +6x+x]}{(4-x^2)^2(6x+1)^{2/3}}\]
Simplify this a bit more and you'll have: \(\boxed{ \dfrac{2[5x^2+x+4]}{(4-x^2)^2(6x+1)^{2/3}}}\)

- Nnesha

sorry i was afk
great explanation jhanny!

- Jhannybean

Thank you :)

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