How do you divide #(x^3 - 7x - 6)/(x+1)#? Algebra Rational Equations and Functions Division of Polynomials 1 Answer Cem Sentin Apr 9, 2018 Quotient is #x^2-x-6# and remainder is #0# Explanation: #x^3-7x-6# =#x^3+x^2-x^2-x-6x-6# =#x^2*(x+1)-x*(x+1)-6*(x+1)# =#(x^2-x-6)*(x+1)# Hence quotient is #x^2-x-6# and remainder is #0# 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 1296 views around the world You can reuse this answer Creative Commons License