Simplify using Factoring after Quotient Rule

AI Thread Summary
The discussion focuses on simplifying a complex derivative expression using factoring after applying the quotient rule in calculus. The original expression involves common factors of (x^2 + 9) and 2x, which can be factored out to simplify the numerator. Participants discuss the steps to factor these common elements correctly and apply the distributive law to combine terms. Clarifications are sought on how certain factors disappear during the simplification process. Overall, the thread emphasizes understanding the algebraic manipulation required for effective simplification.
Econguy
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I am taking an online Introductory Calculus course. I have a decent understanding thus far, however, the problem I'm working on gets somewhat messy and I am having a difficult time simplyifing the answer.

f"(x) = (x^2 + 9)^2 (-2x) - [(9 - x^2)(2)(x^2 + 9)(2x)]/(x^2 + 9)^4

the solutions manual I have been given gives the following simplified answer:

= (2x)(x^2 + 9) [-(x^2 + 9) -2(9 - x^2)]/(x^2 +9)^4

= 2x(x^2 - 27)/(x^2 + 9)^3

I am having a difficult time understanding the factoring involved in the algebraic simplification. I understand how to factor by grouping, trinomials etc. but I just can't seem to understand this. If someone could help fill in the missing steps that my solutions manual doesn't include it would be a great help. Even providing an explanation on the steps involved would help me understand the process much better.

Thanks
 
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In the numerator, (x^2 + 9)^2 (-2x) - (9 - x^2)(2)(x^2 + 9)(2x), the two terms have common factors of x^2 + 9 and 2x. If you take these common factors out of the first term, you have 2x(x^2 + 9)[(x^2 + 9)(-1)]. Do the same thing to the second term.
 
Mark44 said:
In the numerator, (x^2 + 9)^2 (-2x) - (9 - x^2)(2)(x^2 + 9)(2x), the two terms have common factors of x^2 + 9 and 2x. If you take these common factors out of the first term, you have 2x(x^2 + 9)[(x^2 + 9)(-1)]. Do the same thing to the second term.

Thanks for the response: Am I on the right track...

2x(x^2 + 9)[(x^2 + 9)(-1)] - [(2x)(x^2 + 9) (18 - 2x^2)]

at this point I'm not clear how the 2x from the second term disappears (i.e is factored out..).
I understand in what I just did nothing has been factored yet, just rewritten. From this point how does the 2x(x^2 + 9) factor out..

thanks
 
Econguy said:
Thanks for the response: Am I on the right track...

2x(x^2 + 9)[(x^2 + 9)(-1)] - [(2x)(x^2 + 9) (18 - 2x^2)]
Now use the idea that a * b + a * c = a(b + c); i.e., the distributive law. Here a = 2x(x^2 + 9)
Econguy said:
at this point I'm not clear how the 2x from the second term disappears (i.e is factored out..).
2x = 2x * 1, so there's an implied factor of 1 remaining in the second term.
Econguy said:
I understand in what I just did nothing has been factored yet, just rewritten. From this point how does the 2x(x^2 + 9) factor out..

thanks
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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