6.

((2x^2-x-15)/(x^2+3x+5))/((2x^2-9x+10)/(x^2-5x+6))

 

This is a division problem, so we need to invert and multiply.

 

((2x^2-x-15)/(x^2+3x+5))((x^2-5x+6)/(2x^2-9x+10))

 

Now we factor everything.

 

((2x+5)(x-3)(x-2)(x-3))/((x^2+3x+5)(2x-5)(x-2))

 

Only the x - 2s cancel. Note that the bottom of the first fraction does not factor. This is not unusual. Most quadratic trinomials do not factor. If an expression does not factor, then don't factor it. The reason we try to factor is to find factors that will cancel. In this case, it doesn't factor and it will not cancel with anything. So we have to just leave it.

 

((2x+5)(x-3)(x-3))/((x^2+3x+5)(2x-5))