How do you divide #(x^4+2x^3-2x^2+9x+3)/(x^2+1) #? Algebra Rational Equations and Functions Division of Polynomials 1 Answer Martin C. Mar 24, 2018 #x^2+2x-3+(7x+6)/(x^2+1)# Explanation: By long division #(x^4+2x^3-2x^2+9x+3):(x^2+1)=x^2# #-x^4-x^2# #(0+2x^3-3x^2+9x+3):(x^2+1)=x^2+2x# #-2x^3-2x# #(0+0-3x^2+7x+3):(x^2+1)=x^2+2x-3# #+3x^2+3# #(0+0+0+7x+6):(x^2+1)=x^2+2x-3# #x^2+2x-3+(7x+6)/(x^2+1)# Answer link Related questions What is an example of long division of polynomials? How do you do long division of polynomials with remainders? How do you divide #9x^2-16# by #3x+4#? How do you divide #\frac{x^2+2x-5}{x}#? How do you divide #\frac{x^2+3x+6}{x+1}#? How do you divide #\frac{x^4-2x}{8x+24}#? How do you divide: #(4x^2-10x-24)# divide by (2x+3)? How do you divide: #5a^2+6a-9# into #25a^4#? How do you simplify #(3m^22 + 27 mn - 12)/(3m)#? How do you simplify #(25-a^2) / (a^2 +a -30)#? See all questions in Division of Polynomials Impact of this question 1415 views around the world You can reuse this answer Creative Commons License