# Factoring involving Fractional exponents

• Feb 2nd 2009, 06:52 PM
xxabbasxx
Factoring involving Fractional exponents
Hey guys, i was wondering if someone cud help me out with this problem!
Thanks in Advance !

Factor and simplify

(x+4)^1/5 - (x+4)^6/5

need some help asap!(Hi)
• Feb 2nd 2009, 07:12 PM
Re :
$\displaystyle (x+4)^{\frac{1}{5}}-(x+4)^{\frac{6}{5}}$

$\displaystyle =(x+4)^{\frac{1}{5}}(1-(x+4)^6)$
• Feb 2nd 2009, 07:14 PM
xxabbasxx
thanks for the reply and the answer... but i didnt understand how you did that, if you can just explain a little , i would really appreciate it , thank you very much (Happy)
• Feb 2nd 2009, 07:19 PM
Take out the common factor which is $\displaystyle (x+4)^{\frac{1}{5}} .$

so the first part of the equation reduces to 1 and the second part has a power of 6 ( Note that $\displaystyle \frac{1}{5}\times6=\frac{6}{5}$)
• Feb 2nd 2009, 07:23 PM
xxabbasxx
Thank you!
• Feb 2nd 2009, 07:30 PM
Jhevon
Quote:

Originally Posted by mathaddict
$\displaystyle (x+4)^{\frac{1}{5}}-(x+4)^{\frac{6}{5}}$

$\displaystyle =(x+4)^{\frac{1}{5}}(1-(x+4)^{\color{red}6})$

actually, the powers must ADD to give the original. so that 6 should be 1, since 1 + 1/5 gives the original 6/5 for the power of the second term.
• Feb 2nd 2009, 07:55 PM
xxabbasxx
Alright.. Thank you, that will be noted
• Feb 2nd 2009, 07:57 PM
Jhevon
Quote:

Originally Posted by xxabbasxx
Alright.. Thank you, that will be noted

yes, it should be noted already that $\displaystyle x^a \cdot x^b = x^{a + b}$