# 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.

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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 $x$.

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Why did you use u^2 why not just u? – user138246 Dec 17 '11 at 22:47
The $u^2$ gets rid of the fractional exponents, as you asked. André Nicolas shows what happens if you don't. Both are correct and arrive at the same point at the end. It is a matter of taste how to get there. – Ross Millikan Dec 17 '11 at 22:50

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.

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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.

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Please see here for a basic tutorial on writing MathJax so that the equations are readable. – user61527 Sep 4 '13 at 3:32