Question 144510
{{{(48x+48)/(10x-4)}}} * {{{(40x-16)/(12x^2-12)}}}
:
we can do some major factoring here:
{{{(48(x+1))/(2(5x-2))}}} * {{{(8(5x-2))/(12(x^2-1))}}}
:
Note that (x^2-1) can be factored as the "difference of squares"
{{{(48(x+1))/(2(5x-2))}}} * {{{(8(5x-2))/(12(x-1)(x+1))}}} 
;
Cancel the numbers first and you have:
{{{(24(x+1))/((5x-2))}}} * {{{(2(5x-2))/(3(x-1)(x+1))}}}
then 3 into 24
 {{{(8(x+1))/((5x-2))}}} * {{{(2(5x-2))/((x-1)(x+1))}}}
:
Cancel (x+1) and (5x-2)
8 * {{{2/((x-1))}}} = {{{16/(x-1)}}}
:
How about this? did it make sense to you? any questions?