2. ## Re: factor completely

I'm having a hardtime reading the image.

Is it

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

yes it is

4. ## Re: factor completely

The image is a bit hard to read, but I presume you have:

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

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

What do you get?

5. ## Re: factor completely

i get 1/4x^-1/2(3x+1)^-1/2
how do you guys write it perfectly??

6. ## Re: factor completely

Originally Posted by mathisfun26
i get 1/4x^-1/2(3x+1)^-1/2
how do you guys write it perfectly??
We use a typesetting language called latex. Below is the way to write a simple equation in latex. Using Latex, you can write math and it is beautiful and easy to read

The below link shows you how to use Latex.
Introducing LaTeX Math Typesetting

7. ## Re: factor completely

Yes, that's what will be out front, and what will be the factor left within parentheses?

We are using $\displaystyle \LaTeX$ to write the math expressions. Do a search here and online to find out how to use it. It is fairly straightforward.

8. ## Re: factor completely

$\displaystyle \frac{1}{4}x^{-1/2}(3x+1)^{-1/2}$
thanks!

9. ## Re: factor completely

Originally Posted by mathisfun26
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{1}{4}x^{-1/2}(3x@plus;1)^{-1/2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{1}{4}x^{-1/2}(3x+1)^{-1/2}" title="\frac{1}{4}x^{-1/2}(3x+1)^{-1/2}" /></a>
When ever you want to type latex start with [tex ] latex stuff [/ tex]. For example [tex ] \frac{1}{2} [/ tex].

10. ## Re: factor completely

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.

11. ## Re: factor completely

$\displaystyle \frac{1}{2} (3x+1)$ this is what i got when i divided it by the first term i'm not sure how to do it after that

12. ## Re: factor completely

You are close...the first term within the parentheses would be:

$\displaystyle 2(3x+1)$

What do you get for the second term?

13. ## Re: factor completely

why would it be 2??

14. ## Re: factor completely

i got -3x for the second

15. ## Re: factor completely

It would be 2 because:

$\displaystyle \frac{\frac{1}{2}}{\frac{1}{4}}=2$

You are correct with the second term being $\displaystyle -3x$. So, put it all together, and what do you have?

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