What is the remainder when you divide 2x^3+7x^2-4x+7 by x^2+2x-1?

Before we answer the question let's compare the method taught for polynomial long divison to regular long division.If we wanted to find the remainder of 71 when divided by 3, we would find the nearest whole number that when multiplied by 3 is closest to 71 (which would be 23 without going over). We then take off 23x3 to leave us with a remainder of 2.This is the reasoning behind the first step of the polynomial problem. We find the term that when multiplying x^2+2x-1 gets closest to our polynomial with the aim of removing the 2x^3 term.(x^2+2x-1) | (2x^3 +7x^2 - 4x + 7) --> we multiply the quadratic on the left by 2x and then take that away from the cubic equation.STEP 1: 2x*(x^2+2x-1) = 2x^3 + 4x^2 - 2x. Taking that away from our original polynomial: (2x^3 + 7x^2 - 4x + 7) - (2x^3 +4x^2 - 2x) = 3x^2 - 2x + 7(x^2+2x-1) | (3x^2 - 2x + 7) --> we multiply the quadratic on the left by 3 and take that away from the quadratic on the right so we can remove the 3x^2 term.STEP 2: 3*(x^2+2x-1) = 3x^2 + 6x - 3. Taking that away from the quadratic remainder we got from STEP 1, (3x^2 - 2x + 7) - (3x^2 + 6x - 3) = -8x + 10Now we see if we can go any further. However, we very quickly see that our divisor is a quadratic and the remainder from STEP 2 is linear. So we are done and -8x + 10 is our remainder.Putting it all together: (2x^3 +7x^2 - 4x + 7) / (x^2+2x-1) = (2x+3) + (-8x+10)/(x^2+2x-1) To prove to yourself this is true, try to put both terms on the right hand side over the same fraction and check it matches the left hand side.