# Exponents.

#### paulmdrdo

##### Active member
$\displaystyle\frac{-9b^2(a+3b)^{m+2}(2b-4c)^{2+m}}{4(3a+9b)^{2m+2}(b^2-2b^2)^{2-m}}$

$\displaystyle-\left[\frac{2(b-2c)^2b}{9(a+3b)}\right]^m$

i used some factorization of some quantity to arrive to this answer. but i'm not sure how did that technique works.

for example in the quantity $(2b-4c)^{2+m}$ i factored out $2^{2+m}$ to that quantity to have $2^{2+m}(b-2c)^{2+m}$ but i don't know what theorem made this step valid can you tell me why this factorization works?

thanks!

#### Reckoner

##### Member
for example in the quantity $(2b-4c)^{2+m}$ i factored out $2^{2+m}$ to that quantity to have $2^{2+m}(b-2c)^{2+m}$ but i don't know what theorem made this step valid can you tell me why this factorization works?
For this you are making use of the following property of exponents: for real numbers $a,$ $b,$ and $n,$ $(ab)^n = a^nb^n.$

So,
\begin{align*}
(2b-4c)^{2+m} &= \big[\color{red}{2}\color{blue}{(b - 2c)}\big]^{2 + m}\\
&= \color{red}{2}^{2 + m}\color{blue}{(b - 2c)}^{2 + m}.
\end{align*}