Rewrite the divisor 3x-13x−1 as 3(x-1/3)3(x−13).
Then the problem becomes
(6x^3+7x^2+12x-5)/(3(x-1/3)6x3+7x2+12x−53(x−13) or ((6x^3+7x^2+12x-5)/(x-1/3))*1/3(6x3+7x2+12x−5x−13)⋅13
Do synthetic division using the divisor (x-1/3)(x−13). Pull down the 6, write it under the line and multiply it by 1/3. The product 2 is written under the next number, 7. Add the 7 and 2. The sum 9 is written below the line. Multiply 9 by 1/3. The product 3 is written under the next number, 12. Add 12 and 3. The sum 15 is written under the line. Multiply 15 by 1/3. The product 5 is written under the next number, -5. Add -5 and 5. The sum zero is written below the line. So you are multiplying then adding, multiplying then adding, etc.
1/3 | 6...7....12...-5
......|.......2....3.....5
......-------------------
........6....9.....15....0
This result can be written as 6x^2+9x+56x2+9x+5
But remember to multiply this result by 1/313!!
1/3(6x^2+9x+15) = 2x^2 +3x+513(6x2+9x+15)=2x2+3x+5
