Question

More review! Help appreciated

Answer 1

4e. Factor the following expression: \[3x ^{\frac{ 3 }{ 2 }} -9x ^{\frac{ 1 }{ 2 }} +6x ^{-\frac{ 1 }{ 2 }}\]

Answer 2

I'm pretty lost on this, having to factor with the rational fractions is throwing me into a bit of a loop.

Answer 3

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?

Answer 4

hold on i gotta write it :)

Answer 5

fair enough, thank you @PinkiePug

Answer 6

\[3(x \frac{ 1 }{ 2 } - 3x \frac{ 1 }{ 2 }+2 x \frac{ 1 }{ 2 }\]
this is what i got

Answer 7

\[x^{1/2} \implies \sqrt{x}\]
\[x^{-1/2} \implies \frac{ 1 }{ \sqrt{x} }\]

This may be easier to visualize, so we can write it as \[3x^{3/2}-9\sqrt{x}+6\frac{ 1 }{ \sqrt{x} }\]

Answer 8

Then we can find the common denominator \[\sqrt{x}\]

Answer 9

Your numerator should simplify into something better which you can factor

Answer 10

Wait. so are you suggesting that we have to get rid of the negative exponents in order to successfully factor this?

Answer 11

I would factor out 3x^-1/2
\[ 3x ^{\frac{ 3 }{ 2 }} -9x ^{\frac{ 1 }{ 2 }} +6x ^{-\frac{ 1 }{ 2 }} \\
3x ^{-\frac{ 1 }{ 2 }}\left( x^2 -3x +2\right)
\]

Answer 12

now you can factor the quadratic

Answer 13

Yeah that works to haha :P

Answer 14

\[\frac{ 3x^2-9x+6 }{ \sqrt{x} }\] is what you get from my suggestion

Answer 15

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)

Answer 16

yes, or put sqr(x) in the denominator
it is still ugly. but it's factored

Answer 17

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?

Answer 18

it is a common trick. no guarantees, but something to try

Answer 19

Fair enough. I'll keep that in mind. Thank you very much kind sir