# How To factor fractional exponents?????????

• Jan 18th 2009, 04:24 AM
Mudd_101
How To factor fractional exponents?????????
Hey, I was having problems factoring/simplifying a problem I was given involving fractional and negative exponents

This is the problem but I think I can figure out how to do it if you show me the general idea of factoring fractional exponents:

-1/2{(2x^2-x+1)^(-1/2)}(4x-1){(x^3+1)^(1/3)} + {(2x^2 -x + 1)^(1/2}(1/3){(x^3)+1}(3x^2)

• Jan 18th 2009, 04:44 AM
earboth
Quote:

Originally Posted by Mudd_101
Hey, I was having problems factoring/simplifying a problem I was given involving fractional and negative exponents

This is the problem but I think I can figure out how to do it if you show me the general idea of factoring fractional exponents:

-1/2{(2x^2-x+1)^(-1/2)}(4x-1){(x^3+1)^(1/3)} + {(2x^2 -x + 1)^(1/2}(1/3){(x^3)+1}(3x^2)

1. The common denominator of the 2 summands is $\displaystyle \sqrt{2x^2-x+1}$

2. From the sum in the numerator you can factor out $\displaystyle (x^3+1)^{\frac13}$

3. Nevertheless the second factor looks like a nightmare :D
• Jan 18th 2009, 05:15 AM
Mudd_101
Hey, uhh.... my math is really not very good so excuse any silly questions I may ask but uhh........ you said that both denominators are (sorry don't know how to put other people's quotes in)

((2x^2 - x +1)^1/2) but one of the ((2x^2 - x +1)^1/2) has a positive exponent, the other negative. I cannot combine two fractions, one with ((2x^2 - x +1)^(-1/2) and the other ((2x^2 - x +1)^(1/2) can I?

tthanks
• Jan 18th 2009, 07:46 AM
Soroban
Hello, Mudd_101!

It looks like you were differentiating a product.

Quote:

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

I assume the function was: .$\displaystyle f(x) \:=\:\left(2x^2 - x + 1\right)^{\frac{1}{2}}\left(x^3+1\right)^{\frac{1} {3}}$

Then: .$\displaystyle f'(x) \;=\;\tfrac{1}{2}\left(2x^2-x+1\right)^{-\frac{1}{2}}(4x-1)\cdot\left(x^3+1\right)^{\frac{1}{3}} \;+\; (2x^2-x+1)^{\frac{1}{2}}\cdot\tfrac{1}{3}\left(x^3+1\rig ht)^{-\frac{2}{3}}(3x^2)$

. . . . . $\displaystyle f'(x)\;= \;\frac{(4x-1)(x^2+1)^{\frac{1}{3}}}{2(2x^2-x+1)^{\frac{1}{2}}} + \frac{x^2(2x^2-x+1)^{\frac{1}{2}}}{(x^3+1)^{\frac{2}{3}}}$

Get a common denominator:

. . $\displaystyle {\color{blue}\frac{(x^3+1)^{\frac{2}{3}}}{(x^3+1)^ {\frac{2}{3}}}} \cdot \frac{(4x-1)(x^3+1)^{\frac{1}{3}}}{2(2x^2-x+1)^{\frac{1}{2}}}\: +$ $\displaystyle {\color{blue}\frac{2(2x^2-x+1)^{\frac{1}{2}}}{2(2x^2-x+1)^{\frac{1}{2}}}}\cdot\frac{x^2(2x^2-x+1)^{\frac{1}{2}}}{(x^3+1)^{\frac{2}{3}}}$

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

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

. . $\displaystyle = \;\frac{8x^4 - 3x^3 + 2x^2 + 4x - 1}{2(x^3+1)^{\frac{2}{3}}(2x^2-x+1)^{\frac{1}{2}}}$

• Jan 18th 2009, 08:16 AM
Mudd_101
Ohhh!! Thanks Soroban I get it now.

I will click that thanks thing for you.

help was very appreciated bye.