Question 172720
(3x^2 - 9x - 3)/(3x -3)
= all can be divided by 3, so
= (x^2-3x-1)/(x-1)
USING REMAINDER THEOREM
x^2-3x-1= (Ax+B)(x-1)+ R
 apply x=1
then -3 =0 + R
       ******************R=-3*****************
We can also find A and B,
when x=0
     -1 = -B -3
      B = -2
     x=2
4-6-1   = (2A-2)(1) -3
  0     =  2A-2
      A = 1
So, 
(3x^2-9x-3)= (3x-3)(x-2)+ (-3)