Jamierox4ev3r
  • Jamierox4ev3r
More review! Help appreciated
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Jamierox4ev3r
  • Jamierox4ev3r
4e. Factor the following expression: \[3x ^{\frac{ 3 }{ 2 }} -9x ^{\frac{ 1 }{ 2 }} +6x ^{-\frac{ 1 }{ 2 }}\]
Jamierox4ev3r
  • Jamierox4ev3r
I'm pretty lost on this, having to factor with the rational fractions is throwing me into a bit of a loop.
Jamierox4ev3r
  • Jamierox4ev3r
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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More answers

anonymous
  • anonymous
hold on i gotta write it :)
Jamierox4ev3r
  • Jamierox4ev3r
fair enough, thank you @PinkiePug
anonymous
  • anonymous
\[3(x \frac{ 1 }{ 2 } - 3x \frac{ 1 }{ 2 }+2 x \frac{ 1 }{ 2 }\] this is what i got
Astrophysics
  • Astrophysics
\[x^{1/2} \implies \sqrt{x}\] \[x^{-1/2} \implies \frac{ 1 }{ \sqrt{x} }\] This may be easier to visualize, so we can write it as \[3x^{3/2}-9\sqrt{x}+6\frac{ 1 }{ \sqrt{x} }\]
Astrophysics
  • Astrophysics
Then we can find the common denominator \[\sqrt{x}\]
Astrophysics
  • Astrophysics
Your numerator should simplify into something better which you can factor
Jamierox4ev3r
  • Jamierox4ev3r
Wait. so are you suggesting that we have to get rid of the negative exponents in order to successfully factor this?
phi
  • 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
  • phi
now you can factor the quadratic
Astrophysics
  • Astrophysics
Yeah that works to haha :P
Astrophysics
  • Astrophysics
\[\frac{ 3x^2-9x+6 }{ \sqrt{x} }\] is what you get from my suggestion
Jamierox4ev3r
  • Jamierox4ev3r
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
  • phi
yes, or put sqr(x) in the denominator it is still ugly. but it's factored
Jamierox4ev3r
  • Jamierox4ev3r
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
  • phi
it is a common trick. no guarantees, but something to try
Jamierox4ev3r
  • Jamierox4ev3r
Fair enough. I'll keep that in mind. Thank you very much kind sir

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