How do you divide #(2x^2-3x)/(3x + 1)#?

To divide (2x^2-3x)/(3x + 1), you can use polynomial long division or synthetic division. Here is the solution using polynomial long division:

Step 1: Divide the first term of the numerator (2x^2) by the first term of the denominator (3x). The result is (2/3)x.

Step 2: Multiply the entire denominator (3x + 1) by the result from step 1, which is (2/3)x. This gives you (2/3)x * (3x + 1) = (2/3)x * 3x + (2/3)x * 1 = 2x^2 + (2/3)x.

Step 3: Subtract the result from step 2 from the numerator (2x^2 - 3x) to get the remainder. (2x^2 - 3x) - (2x^2 + (2/3)x) = - (11/3)x.

Step 4: Bring down the next term from the numerator, which is - (11/3)x.

Step 5: Divide the term you brought down (- (11/3)x) by the first term of the denominator (3x). The result is - (11/9).

Step 6: Multiply the entire denominator (3x + 1) by the result from step 5, which is - (11/9). This gives you - (11/9) * (3x + 1) = - (11/9) * 3x - (11/9) * 1 = - (11/3)x - (11/9).

Step 7: Subtract the result from step 6 from the remainder (- (11/3)x) to get the new remainder. (- (11/3)x) - (- (11/3)x - (11/9)) = 0.

Step 8: Since the new remainder is 0, the division is complete. The quotient is (2/3)x - (11/9).

Therefore, (2x^2-3x)/(3x + 1) = (2/3)x - (11/9).