# Thread: Factoring to reduce Questions

1. ## Factoring to reduce Questions

In the expression
$\displaystyle \frac{16x^3-24x^2}{8x^4-12x^3}=\frac{8x^2(2x-3)}{4x^3(2x-3)}=\frac{2}{1} * \frac{x^2}{x^3}= \frac{2}{x}$

I understand everything but the last step, how is the $\displaystyle \frac{2}{1} * \frac{x^2}{x^3}= \frac{2}{x}$?

If it were me I would think the correct thing to do in this case would be to apply the exponent property$\displaystyle \frac{a^m}{a^n}=a^{m-n}$, thus equaling $\displaystyle 2x^{2-3} =\frac{1}{2x}$

Thanks!

2. Originally Posted by allyourbass2212
In the expression
$\displaystyle \frac{16x^3-24x^2}{8x^4-12x^3}=\frac{8x^2(2x-3)}{4x^3(2x-3)}=\frac{2}{1} * \frac{x^2}{x^3}= \frac{2}{x}$

I understand everything but the last step, how is the $\displaystyle \frac{2}{1} * \frac{x^2}{x^3}= \frac{2}{x}$?

If it were me I would think the correct thing to do in this case would be to apply the exponent property$\displaystyle \frac{a^m}{a^n}=a^{m-n}$, thus equaling $\displaystyle 2x^{2-3} =\frac{1}{2x}$

Thanks!
$\displaystyle \frac 21 \cdot \frac {x^2}{x^3} = \frac 21 \cdot x^{2 - 3} = \frac 21 \cdot \frac 1x = \frac 2x$

the 2 is NOT affected by the power you have here. this is why it is important for you to get into the habit of using parentheses properly.

if it were $\displaystyle (2x)^{2-3}$ you would be correct. that is $\displaystyle \frac 1{2x}$. but in $\displaystyle 2x^{2-3}$, only the $\displaystyle x$ is affected by the power. the answer is $\displaystyle \frac 2x$