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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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3 Answers

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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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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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