How to decompose and resolve respect to x

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To decompose and resolve the expression \frac{a}{b\pm x}, it's important to note that it represents two separate fractions: \frac{a}{b+x} and \frac{a}{b-x}. A space is required between "\pm" and "x" to ensure proper LaTeX formatting, as the absence of this space leads to errors. The term "decompose with respect to x" needs clarification, as neither fraction can be further simplified. Understanding these points is crucial for accurate mathematical representation and manipulation.
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How to decompose and resolve respect to x\frac{a}{b\pmx}?
 
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I would say a/b+x.
 


Ignition said:
How to decompose and resolve respect to x\frac{a}{b\pm x}?
You need a space between "\pm" and "x". The way you have it, LaTex tries to find a command "\pmx" and since there isn't one, it shows nothing.
\frac{a}{b\pm x} means, of course, two values: \frac{a}{b+ x} and \frac{a}{b- x}. Now, what do you mean by "decompose with respect to x"? Neither of those fractions can be simplified more.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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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