Simplifying this expression:

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January 10th 2009, 09:15 PM
Kaln0s

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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. (Headbang)

Thanks for any help guys.
January 10th 2009, 09:54 PM
Grandad

Simplifying

Hello Kaln0s

Quote:

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. (Headbang)

Thanks for any help guys.

Write the expression as:

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

Then write it all over a common denominator:

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

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

Grandad
January 10th 2009, 10:13 PM
Kaln0s

Quote:

Originally Posted by Grandad

Hello Kaln0s
Write the expression as:

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

Then write it all over a common denominator:

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

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

Grandad

What happened to the 1/4? Out of curiosity. I'm still kindof confused? Thanks for the help though. ^_^;
January 10th 2009, 10:29 PM
Grandad

Simplifying

Hello Kaln0s

Quote:

Originally Posted by Kaln0s

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

$6x(2x+1)^\frac{1}{4} (x^2+1)^2 = \frac{6x(2x+1)^\frac{1}{4} (x^2+1)^2 \times (2x+1)^\frac{3}{4}}{(2x+1)^\frac{3}{4}}$

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

Can you see that OK?

Grandad