Factoring with fractional exponents I really hate to keep asking questions but I just can't figure this out, I don't know what is wrong with me but I can't figure it out. I stared at it for 5 minutes and not a thought came into my head on how to do it that actually accomplished anything.
$3x^{3/2}-9x^{1/2}+6x^{-1/2}$ I am pretty sure I can't factor this with crazy exponents but I don't know how to get rid of them and keep the problem the same. At least in any way that simplifies things.
 A: Try replacing $x$ with $u^2$ to clear the fractional exponents.
If $u^2=x$, we get $3x^{3/2}-9x^{1/2}+6x^{-1/2} = 3(u^2)^{3/2}-9(u^2)^{1/2}+6(u^2)^{-1/2}$ 
$= 3u^3-9u+6u^{-1}$
Next, factor out $u^{-1}$ and get $3u^{-1}(u^4-3u^2+2)$. 
This is now quadratic in $u^2$ (which is $x$) so  we get $3x^{-1/2}(x^2-3x+2) = $
$$ \frac{3(x-2)(x-1)}{\sqrt{x}}$$
Of course, you can avoid introducing $u$ if you see that factoring out $x^{-1/2}$ at the beginning leaves you with a quadratic in $x$.
A: How one should "simplify," if at all, depends on the problem we are trying to solve.  Let
$$f(x)=3x^{3/2}-9x^{1/2}+6x^{-1/2}.$$
We want to make this look nicer.  The fractional exponents are unpleasant.  We can get rid of them all by multiplying through by $x^{1/2}$.  But then to keep $f(x)$ unchanged, we will need to divide by $x^{1/2}$.
Now we carry out the strategy:
$$f(x)=\frac{x^{1/2}(3x^{3/2}-9x^{1/2}+6x^{-1/2})}{x^{1/2}}=\frac{3x^2-9x+6}{x^{1/2}}.$$
The top factors as $3(x-1)(x-2)$, and we conclude that 
$$f(x)=\frac{3(x-1)(x-2)}{x^{1/2}}.\qquad\qquad (\ast)$$
For some purposes, this is more useful that the original expression for $f(x)$.
For example, if we need to know where $f(x)=0$, we can read off from $(\ast)$ that the roots are $x=1$ and $x=2$.  But the original form with the fractional exponents may be the more useful one for other purposes.
A: The question is asking for this answer: 3x^(-1/2)(x-1)(x+2) 
It assumes you can start out by factoring out 3x^(-1/2), so you have 3x^(-1/2)(y^2 -3y +2), which factors into the answer. 
