Question 232377
Whenever you are factoring, keep an eye out for factoring out a -1. It can always be done and sometimes doing so makes a problem easier. Since factoring trinomials is easier with a positive leading coefficient, I am going to start by factoring out -1:
{{{(-x^2+10x-21)/(-3x^2+23y-14)}}}
{{{((-1)(x^2-10x+21))/((-1)(3x^2-23y+14))}}}
The trinomials are now easier to factor, especially the first one. The second one is a bit more difficult. Since 3 factors in only one way, 3 = 3*1, the factors will look like:
(3x    )(x   )
if it factors at all. And since the 14 is positive and the 23 is negative, the factors will look like:
(3x -  )(x - )
Now we just have to figure out which factors of 14 (1 and 14 or 2 and 7) to use and where to place them in the above so that we would end up with 23 in the middle if you we to multiply.
{{{((-1)(x-7)(x-3))/((-1)(3x - 2)(x-7))}}}
Now we can cancel common factors:
{{{(cross((-1))cross((x-7))(x-3))/(cross((-1))(3x - 2)cross((x-7)))}}}
leaving
{{{(x-3)/(3x-2)}}}
(NOTE: Do not cancel the x's or the 3's here. They are <b>not</b> factors!)