# simplify $3x^{3/2}-9^{1/2}+6^{-1/2}$

This problem has been posted before, unfortunately I still can't understand it from the other example. What I keep getting stuck on is the middle value and factoring out the negative fractional exponent. the problem goes

$3x^{3/2}-9x^{1/2}+6x^{-1/2}$

I know I need to factor out $3x^{-1/2}$ the problem is I can't understand how to do that from the middle monomial $9x^{1/2}$ when it doesn't have the negative value.

• Are you missing some $x$s? Jan 31 '14 at 4:18

If you aren't missing $x$s, then just simplify $9^{1/2}+6^{-1/2}$. If you are missing $x$s, factor out the lowest exponent: $x^{-1/2}(3x^2 - 9x + 6) = 3x^{-1/2}(x^2 - 3x + 2)$

Edit: You can think of $9x^{1/2}$ as $9x^1 \times x^{-1/2}$. This works because $x^a \times x^b = x^{ab}$.

From there you can see how I factored out $x^{-1/2}$.

• I was missing "x's" but my problem is how am I pulling out a $-(1/2)$ from the $9^(1/2)$ if it doesn't have a $-(1/2)$ Jan 31 '14 at 12:31

It is known that $a^m\cdot{a^n}=a^{m+n}$. You can write $\frac{3}{2}=2-\frac{1}{2} \ , \ \frac{1}{2}=1-\frac{1}{2}$, thus $$3x^{1.5}-9x^{0.5}+6x^{-0.5}=3x^{-0.5}[x^2-3x+2]=3x^{-0.5}(x-1)(x-2)$$

I hope it helps.

• it did, I finally understand it. thank you. Jan 31 '14 at 14:29