X^3/2 - 2x^-1/2 + x^1/2 Help Factoring

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The discussion revolves around factoring the expression X^3/2 - 2x^-1/2 + x^1/2. Participants express confusion over the solution provided in the textbook, which is 1 - x^4/3, and share their own attempts, such as factoring out X^-3/2. There is a suggestion to use substitution, specifically u = x^{1/2}, to simplify the factoring process. Some users admit to posting in the wrong section and acknowledge their mistakes regarding the answer key. The overall focus remains on understanding the correct method to factor the expression.
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I'm looking to factor this...


X^3/2 - 2x^-1/2 + x^1/2

The textbook offers the solution 1-x ^4/3

No idea how to get this result. I got X ^-3/2 (x-1)^2
I would like it if someone could shed light on how to factor x^-3/2 from things like 2x ^1/2
I assume you just keep the equation balance, so if you re-distribute everything, it reverts to the original.

I would like to know how you get the answer too.

Thanks so much!
 
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Maybe you can do substitution, say ##u = x^{1/2}##?
 
Hierophant said:
I'm looking to factor this... X^3/2 - 2x^-1/2 + x^1/2

The textbook offers the solution 1-x ^4/3

No idea how to get this result. I got X ^-3/2 (x-1)^2

Both look wrong
Mod note: Removed complete answers.
 
Last edited by a moderator:
Hierophant said:
I'm looking to factor this...

X^3/2 - 2x^-1/2 + x^1/2

The textbook offers the solution 1-x ^4/3

No idea how to get this result. I got X ^-3/2 (x-1)^2
I would like it if someone could shed light on how to factor x^-3/2 from things like 2x ^1/2
I assume you just keep the equation balance, so if you re-distribute everything, it reverts to the original.

I would like to know how you get the answer too.

Thanks so much!

So why are you looking to factor this, what exactly are you trying to obtain?

Also this should go in homework help, fyi.

edit:
Pwan just gave your answer, but is this what you're even looking for?
 
Last edited:
To my delight and perhaps also not to my delight, I got the correct answer, but used the wrong answer key. Sorry guys, won`t happen again.

Also sorry for posting in the wrong section, I was rather fatigued at the moment of posting.

I`m just going to go sit in the corner now.
 
To my delight and perhaps also not to my delight, I got the correct answer, but used the wrong answer key. Sorry guys, won`t happen again.

Also sorry for posting in the wrong section, I was rather fatigued at the moment of posting.

I`m just going to go sit in the corner now.
 
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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