this is a fraction (x-2)^2 over 4x^2-16 divided by 21x over 3x+6 how do you divide it and make it into its lowest terms?
umm, yes, that's what i have
$\displaystyle \frac {\frac {(x - 2)^2}{4x^2 - 16}}{\frac {21x}{3x + 6}} = \frac {(x - 2)^2}{4x^2 - 16} \div \frac {21x}{3x + 6}$
$\displaystyle = \frac {(x - 2)^2}{4x^2 - 16} \cdot \frac {3x + 6}{21x}$
$\displaystyle = \frac {(x - 2)^2}{4 \left( x^2 - 4 \right)} \cdot \frac {3(x + 2)}{21x}$
$\displaystyle = \frac {(x - 2)^2}{4(x - 2)(x + 2)} \cdot \frac {3(x + 2)}{21x}$
Hopefully the rest is obvious to you. if not, it's ok, say so