Question 56357
{{{ ((x^2+2x)/(4x-3))/(6x^3/(8x^2-2x-3)) }}}
there is a property that says
{{{ (a/b)/(c/d) = ad/bc }}}
So lets re write the complex fraction into a simpler one
{{{((x^2+2x)(8x^2-2x-3))/((4x-3)(6x^3)) }}}
Factor the 8x^2-2x-3 ....
{{{((x^2+2x)(4x-3)(2x-2))/((4x-3)(6x^3)) }}}
cancel the 4x -3 on top and bottom
{{{((x^2+2x)(2x-2))/(6x^3) }}}
factor out an x from the first quantity and a 2 out of the second
{{{(2x(x+2)(x-1))/(6x^3) }}}
Reduce the 2x and the 6x^3
{{{((x+2)(x-1))/(3x^2) }}}
I would accept this answer ... your teacher may want you to foil out the numerator, in which case it would look like this
{{{ (x^2+1x -2)/(3x^2) }}}