$\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?

Printable View

- Nov 24th 2012, 07:47 PMmathisfun26Re: 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 PMmathisfun26Re: factor completely
$\displaystyle \frac{3x+2}{4^{x^1/2} (3x+1)^{1/2}} $ would it be this?

- Nov 24th 2012, 08:09 PMMarkFLRe: 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 PMmathisfun26Re: factor completely
$\displaystyle \frac{1}{4}x^{-1/2}(3x+1)^{-1/2}[1] $ ??

- Nov 24th 2012, 08:53 PMMarkFLRe: 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 PMmathisfun26Re: factor completely
[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 PMMarkFLRe: factor completely
$\displaystyle \frac{1}{2}=\frac{1}{4}\cdot2$ and $\displaystyle \frac{3}{4}=\frac{1}{4}\cdot3$