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How do you factor $3x^{3/2} -9x^{1/2}+6x^{-1/2}$ ?

I factored out a 3 to get:

$3(x^{3/2} -3x^{1/2}+2x^{-1/2})$, but it seems this can be factored further.

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    $\begingroup$ Factor out $3x^{-1/2}$ first, not just $3$. $\endgroup$ – Thomas Andrews Sep 1 '15 at 22:54
  • $\begingroup$ It might help if you let a new variable $y=x^{-1/2}$. $\endgroup$ – Mark Viola Sep 1 '15 at 23:04
  • $\begingroup$ No, that really doesn't help, @Dr.MV Something like it might help. $\endgroup$ – Thomas Andrews Sep 2 '15 at 1:05
  • $\begingroup$ @ThomasAndrews I can see a way that it might help. $\endgroup$ – Mark Viola Sep 2 '15 at 1:40
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$$ 3 x^{\frac{3}{2}} -9 x^{\frac{1}{2}} + 6x^{-\frac{1}{2}}$$

Factor out $\sqrt{x}$ to get

$$\sqrt{x} (3 x - 9 + 6x^{-1})$$

Furthermore you get

$$\sqrt{x} 3(x - 3 + 2x^{-1})$$

and then

$$\sqrt{x} 3 \frac{(x-2)(x-1)}{x}$$

Simplify:

$$3 \frac{(x-2)(x-1)}{\sqrt{x}}$$

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Using Thomas Andrews' suggestion, you get

$\displaystyle3x^{3/2}-9x^{1/2}+6x^{-1/2}=3x^{-1/2}\big(x^2-3x+2\big)=3x^{-1/2}(x-1)(x-2)=\frac{3(x-1)(x-2)}{x^{1/2}}$

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