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I just found the derivative of a function and now I'm trying to factor and simplify the expression below. I have no idea how to factor terms with fractions as exponents especially when these fractions are expressed as a negative exponent. Could someone help me.

Also I don't have enough reps to post pictures. I would really appreciate it if yall could bump my rep points up so I can construct my questions more clearly in the future.

The equation is as follows.....

$-45x^3 (9x^2 + 3)^{-3/2} + 10x (9x^2 + 3)^{-1/2}$

It would be great if someone could explain in detail "clearly" how to go about factoring an expression with negative fractions as exponents. In other words I need clarification on the process instead of the answer.

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  • $\begingroup$ The process of factoring an expression which has negative exponents is the same as for positive exponents, it's just about recognising things like in arbautjc's answer. This will become easier with practice and experience. There is a caveat with negative exponents, though, you must be careful when introducing a negative exponent that the object which you are "exponent-iating" (the stuff in the brackets) is non-zero. If you're asking about why we can factorise, well, then that is a different question altogether. The answer is to do with associativity, distributivity and sometimes commutivity $\endgroup$ – user27182 Mar 22 '13 at 19:55
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$$-45x^3 (9x^2 + 3)^{-3/2} + 10x (9x^2 + 3)^{-1/2}=$$

$$=5x(-9x^2) (9x^2 + 3)^{-1/2-1} + 5x\cdot2 (9x^2 + 3)^{-1/2}=$$

$$=5x(-9x^2) (9x^2 + 3)^{-1/2}(9x^2 + 3)^{-1} + 5x\cdot2 (9x^2 + 3)^{-1/2}=$$

$$=5x(9x^2 + 3)^{-1/2}(-9x^2(9x^2 + 3)^{-1} + 2)=$$

$$=5x(9x^2 + 3)^{-1/2}(2-9x^2(9x^2 + 3)^{-1})$$

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Hint: $(9x^2 + 3)^{-1/2}=(9x^2 + 3) \cdot (9x^2 + 3)^{-3/2}$

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Hint: consider the two terms as fractions (negative exponent means that this factor belongs to the denominator...)... make sure the denominator in the two terms is $(9x^2+3)^{3/2}$ Then factorize the fraction...

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  • $\begingroup$ So i can throw the two expressions in () with the negative exponent in the denominator which makes the exponents now positive. I got this. So next when i do factor it I pull out the smallest exponent correct ??? $\endgroup$ – Shane Yost Mar 24 '13 at 0:53

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