2x^2 - 3x + 2 I hope that the formatting isn't destroyed since this is an exercise in long division of polynomials. We have: (2x^3 - 9x^2 + 11x - 6)/(x-3) Look at the highest term in the numerator which is 2x^3. Now divide by the highest term in the divisor which is x. Perform that division, so 2x^3/x = 2x^2. And finally, multiply the divisor by the quotient term you just made and then subtract that from the numerator. So we have Partial quotient = 2x^3/x = 2x^2 Multiply the divisor = (x-3)*2x^2 = 2x^3 - 6x^2 Subtract from numerator = (2x^3 - 9x^2 + 11x - 6) - (2x^3 - 6x^2) = (- 9x^2 + 11x - 6) + 6x^2 = (- 3x^2 + 11x - 6) So we now have a quotient of 2x^2 and a simpler problem of (- 3x^2 + 11x - 6)/(x-3). Let's continue. Partial quotient = -3x^2/x = -3x Multiply the divisor = -3x(x-3) = -3x^2 + 9x Subtract from numerator = (- 3x^2 + 11x - 6) - (-3x^2 + 9x) = (11x - 6) - 9x = 2x - 6 And our quotient so far is now 2x^2 - 3x, and the simpler problem is: (2x - 6)/(x - 3). So let's continue: Partial quotient = 2x/x = 2 Multiply the divisor = 2(x-3) = 2x - 6 Subtract from numerator = (2x - 6) - (2x - 6) = 0 And we're done with the quotient of 2x^2 - 3x + 2
