How do you divide #(x^2-3) / (-6x^4+2x^2-8x+1)#? Algebra Rational Equations and Functions Division of Polynomials 1 Answer Tamir E. Dec 26, 2017 #-2(3x^2+8)-{8x+47}/{-6x^4+2x^2-8x+1}# Explanation: #-6x^2-16# #bar(-6x^4+2x^2-8x+1)|x^2-3# #-# #-6x^4-18x^2# #bar(0x^4-16x^2-8x+1)# #-# #0x^4-16x^2+0x+48# #bar(0x^4-0x^2-8x-47)# 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 1452 views around the world You can reuse this answer Creative Commons License