How do you divide #(x^2)/(2x + 4)#? Precalculus Real Zeros of Polynomials Long Division of Polynomials 1 Answer Konstantinos Michailidis Feb 27, 2016 It is #1/2*(x^2/(x+2))= x/2+2/(x+2)-1# Explanation: It is #1/2*[x^2/(x+2)]=1/2*[(x^2+2x-2x)/(x+2)]=1/2*[x-(2x)/(x+2)]= 1/2*[x-2*(x+2-2)/(x+2)]=1/2*[x-2*(1-2/(x+2))]= x/2-1+2/(x+2)# Answer link Related questions What is long division of polynomials? How do I find a quotient using long division of polynomials? What are some examples of long division with polynomials? How do I divide polynomials by using long division? How do I use long division to simplify #(2x^3+4x^2-5)/(x+3)#? How do I use long division to simplify #(x^3-4x^2+2x+5)/(x-2)#? How do I use long division to simplify #(2x^3-4x+7x^2+7)/(x^2+2x-1)#? How do I use long division to simplify #(4x^3-2x^2-3)/(2x^2-1)#? How do I use long division to simplify #(3x^3+4x+11)/(x^2-3x+2)#? How do I use long division to simplify #(12x^3-11x^2+9x+18)/(4x+3)#? See all questions in Long Division of Polynomials Impact of this question 1247 views around the world You can reuse this answer Creative Commons License