0
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ Are you missing some $x$s? $\endgroup$ Jan 31 '14 at 4:18
2
$\begingroup$

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 $x^{-1/2}$.

$\endgroup$
1
  • $\begingroup$ 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 $-(1/2)$ $\endgroup$
    – Joshhw
    Jan 31 '14 at 12:31
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ it did, I finally understand it. thank you. $\endgroup$
    – Joshhw
    Jan 31 '14 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.