More review! Help appreciated

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4e. Factor the following expression: \[3x ^{\frac{ 3 }{ 2 }} -9x ^{\frac{ 1 }{ 2 }} +6x ^{-\frac{ 1 }{ 2 }}\]
I'm pretty lost on this, having to factor with the rational fractions is throwing me into a bit of a loop.
Does anyone have a simple method of doing this by hand? Or do you recommend grabbing a calculator to look for x-intercepts so I can factor that way?

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hold on i gotta write it :)
fair enough, thank you @PinkiePug
\[3(x \frac{ 1 }{ 2 } - 3x \frac{ 1 }{ 2 }+2 x \frac{ 1 }{ 2 }\] this is what i got
\[x^{1/2} \implies \sqrt{x}\] \[x^{-1/2} \implies \frac{ 1 }{ \sqrt{x} }\] This may be easier to visualize, so we can write it as \[3x^{3/2}-9\sqrt{x}+6\frac{ 1 }{ \sqrt{x} }\]
Then we can find the common denominator \[\sqrt{x}\]
Your numerator should simplify into something better which you can factor
Wait. so are you suggesting that we have to get rid of the negative exponents in order to successfully factor this?
  • phi
I would factor out 3x^-1/2 \[ 3x ^{\frac{ 3 }{ 2 }} -9x ^{\frac{ 1 }{ 2 }} +6x ^{-\frac{ 1 }{ 2 }} \\ 3x ^{-\frac{ 1 }{ 2 }}\left( x^2 -3x +2\right) \]
  • phi
now you can factor the quadratic
Yeah that works to haha :P
\[\frac{ 3x^2-9x+6 }{ \sqrt{x} }\] is what you get from my suggestion
oh. so i can factor \(x^{2}-3x+2\), since that's the quadratic. Oh wow that's much simpler. So the final answer would be: \(3x^{-\Large\frac{1}{2}}\)(x-1)(x-2)
  • phi
yes, or put sqr(x) in the denominator it is still ugly. but it's factored
so just so I know, for future problems like this, is it always acceptable to take out a value so that you're left with a quadratic that you can easily factor?
  • phi
it is a common trick. no guarantees, but something to try
Fair enough. I'll keep that in mind. Thank you very much kind sir

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