Math Help - factor completely

1. Re: factor completely

$\frac{1}{4}x^{-1/2}(3x+1)^{-1/2}[3x+2]$ if the question said no negative exponents in the final answer how would it be?

2. Re: factor completely

$\frac{3x+2}{4^{x^1/2} (3x+1)^{1/2}}$ would it be this?

3. Re: factor completely

You would have:

$\frac{1}{4}x^{-\frac{1}{2}}(3x+1)^{-\frac{1}{2}}\left((3x+1)-3x \right)$

Now simplify...what do you get?

4. Re: factor completely

$\frac{1}{4}x^{-1/2}(3x+1)^{-1/2}[1]$ ??

5. Re: factor completely

Yes, and you may omit the factor of 1. You may choose to write this as:

$\frac{1}{4\sqrt{x(3x+1)}}$

6. Re: factor completely

Originally Posted by MarkFL2
A neat feature here is that if you hover your mouse cursor over an expression written in LaTeX code, you will see the code used to generate the expression.
Originally Posted by jakncoke
When ever you want to type latex start with [tex ] latex stuff [/ tex]. For example [tex ] \frac{1}{2} [/ tex].
[QUOTE=MarkFL2;754391]The image is a bit hard to read, but I presume you have:

$\frac{1}{2}x^{-\frac{1}{2}}(3x+1)^{\frac{1}{2}}-\frac{3}{4}x^{ \frac{1}{2}}(3x+1)^{-\frac{1}{2}}$

First, look at the constant factors...they have $\frac{1}{4}$ as factors. Then look at the variable factors and pull out the factors with the smallest exponents.

quick question how did u get 1/4 from the factors of 1/2 and 3/4

7. Re: factor completely

$\frac{1}{2}=\frac{1}{4}\cdot2$ and $\frac{3}{4}=\frac{1}{4}\cdot3$

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