Question 72267
{{{((6y^2-5y+1)/(12y^2-5y-2))*((4y^2+3y-1)/(3y^2-4y+1))}}}
First factor everything you can, I'm using the ac, arch, or grouping method.  You Replace the middle coefficient with two integers that multply to get the product of the first and last coefficient, but add to get the middle, then you factor by grouping:
{{{((6y^2-3y-2y+1)/(12y^2-8y+3y-2))*((4y^2+4y-y-1)/(3y^2-3y-y+1))}}}
{{{((3y(2y-1)-1(2y-1))/(4y(3y-2)+1(3y-2)))*((4y(y+1)-1(y+1))/(3y(y-1)-1(y-1)))}}}
{{{((3y-1)(2y-1)/((4y+1)(3y-2)))*((4y-1)(y+1)/((3y-1)(y-1)))}}}
{{{((3y-1)(2y-1)(4y-1)(y+1))/((4y+1)(3y-2)(3y-1)(y-1))}}}
{{{(cross(3y-1)*(2y-1)(4y-1)(y+1))/((4y+1)(3y-2)*cross(3y-1)(y-1))}}}
{{{highlight(((2y-1)(4y-1)(y+1))/((4y+1)(3y-2)(y-1)))}}}
You can multiply the top and the bottom out, but teachers generally want you to leave it in factored form.
Happy Calculating!!!