1. ## Simplifying this expression:

Hey guys, I'm having a hard time trying to figure out how to simplify this:

*Problem is attached* If you can't use that it's here:
6x(2x+1)^1/4 (x^2+1)^2 + .5(2x+1)^-3/4 (x^2+1)^3

I need a start, and I'm assuming it's with the exponents but I can't figure it out for my life.

Thanks for any help guys.

2. ## Simplifying

Hello Kaln0s
Originally Posted by Kaln0s
Hey guys, I'm having a hard time trying to figure out how to simplify this:

*Problem is attached* If you can't use that it's here:
6x(2x+1)^1/4 (x^2+1)^2 + .5(2x+1)^-3/4 (x^2+1)^3

I need a start, and I'm assuming it's with the exponents but I can't figure it out for my life.

Thanks for any help guys.
Write the expression as:

$\displaystyle 6x(2x+1)^\frac{1}{4} (x^2+1)^2 + \frac{.5 (x^2+1)^3}{(2x+1)^\frac{3}{4}}$

Then write it all over a common denominator:

$\displaystyle \frac{6x(2x+1) (x^2+1)^2 + .5 (x^2+1)^3}{(2x+1)^\frac{3}{4}}$ (Can you see how I've done that?)

Don't multiply out the brackets - instead, take out a common factor $\displaystyle (x^2+1)^2$. Can you take it from here?

Hello Kaln0s
Write the expression as:

$\displaystyle 6x(2x+1)^\frac{1}{4} (x^2+1)^2 + \frac{.5 (x^2+1)^3}{(2x+1)^\frac{3}{4}}$

Then write it all over a common denominator:

$\displaystyle \frac{6x(2x+1) (x^2+1)^2 + .5 (x^2+1)^3}{(2x+1)^\frac{3}{4}}$ (Can you see how I've done that?)

Don't multiply out the brackets - instead, take out a common factor $\displaystyle (x^2+1)^2$. Can you take it from here?

What happened to the 1/4? Out of curiosity. I'm still kindof confused? Thanks for the help though. ^_^;

4. ## Simplifying

Hello Kaln0s
Originally Posted by Kaln0s
What happened to the 1/4? Out of curiosity. I'm still kindof confused? Thanks for the help though. ^_^;
$\displaystyle 6x(2x+1)^\frac{1}{4} (x^2+1)^2 = \frac{6x(2x+1)^\frac{1}{4} (x^2+1)^2 \times (2x+1)^\frac{3}{4}}{(2x+1)^\frac{3}{4}}$

Then use the fact that $\displaystyle (2x+1)^\frac{1}{4} \times (2x+1)^\frac{3}{4} = (2x+1)^{\frac{1}{4}+\frac{3}{4}} = (2x+1)^1$

Can you see that OK?