How to write an algebraic fraction in a given form e.g. (3+13x-6x^2)/(2x-3) as Ax + B + C/(2x-3) where A, B and C are natural numbers

I would break this down into simple steps:

Step 1 - Equate the 2 expressions: (3 + 13x - 6x²) / (2x - 3) = Ax + B + C / (2x - 3)

Step 2 - Multiply both sides by the demoninator: 3 + 13x - 6x² = (2x - 3)*Ax + (2x - 3)*B + (2x - 3)*C / (2x - 3)

Step 3 - Expand and simplify both sides: 3 + 13x - 6x² = 2Ax² - 3Ax + 2Bx - 3B + C 

Step 4 - Compare the coefficients: -6x² = 2Ax², 13x = (2B - 3A)x, 3 = C - 3B

Step 5 - Solve the coefficients, starting with the simpliest first: -6x² = 2Ax², then -6 = 2A, so -3 = A.

Using A = -3, 13x = (2B - 3(-3))x = (2B + 9)x, then 13 = 2B + 9, hence 4 = 2B, and 2 = B.

Using B = 2, 3 = C - 3(2) = C - 6, then C = 6 + 3 = 9.

So your answer is A = -3, B = 2 and C = 9