How do you divide #(6x^3 + 10x^2 + x + 8) / (2x^2 + 1)#? Algebra Rational Equations and Functions Division of Polynomials 1 Answer Alan P. Nov 14, 2015 #(6x^3+10x^2+x+8)/(2x^2+1) = 3x+5# with remainder #(-2x+3)# Explanation: Using polynomial long division: #{: (,,3x,+5,,), (,,"----","----","----","----"), (2x^2+1,")",6x^3,+10x^2,+x,+8), (,,6x^3,,+3x,), (,,"----","----","----","----"), (,,,10x^2,-2x,+8), (,,,10x^2,,+5), (,,,"----","----","----"), (,,,,-2x,+3) :}# 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 7807 views around the world You can reuse this answer Creative Commons License