factor completely

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• Nov 24th 2012, 07:47 PM
mathisfun26
Re: factor completely
$\displaystyle \frac{1}{4}x^{-1/2}(3x+1)^{-1/2}[3x+2]$ if the question said no negative exponents in the final answer how would it be?
• Nov 24th 2012, 07:52 PM
mathisfun26
Re: factor completely
$\displaystyle \frac{3x+2}{4^{x^1/2} (3x+1)^{1/2}}$ would it be this?
• Nov 24th 2012, 08:09 PM
MarkFL
Re: factor completely
You would have:

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

Now simplify...what do you get?
• Nov 24th 2012, 08:10 PM
mathisfun26
Re: factor completely
$\displaystyle \frac{1}{4}x^{-1/2}(3x+1)^{-1/2}[1]$ ??
• Nov 24th 2012, 08:53 PM
MarkFL
Re: factor completely
Yes, and you may omit the factor of 1. You may choose to write this as:

$\displaystyle \frac{1}{4\sqrt{x(3x+1)}}$
• Nov 24th 2012, 09:05 PM
mathisfun26
Re: factor completely
Quote:

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.

Quote:

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:

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

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.

quick question how did u get 1/4 from the factors of 1/2 and 3/4
• Nov 24th 2012, 09:29 PM
MarkFL
Re: factor completely
$\displaystyle \frac{1}{2}=\frac{1}{4}\cdot2$ and $\displaystyle \frac{3}{4}=\frac{1}{4}\cdot3$
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