How do you divide #( x^3+4x^2+x-6 )/(x-3)#? Algebra Rational Equations and Functions Division of Polynomials 1 Answer Johnson Z. Dec 16, 2015 #((x^2+1)(x+4)(x-6))/(x-3)# Explanation: Assuming that you mean simplifying: #(x^3+4x^2+x-6)/(x-3)# #=(x^2(x+4)+1(x-6))/(x-3)# #=((x^2+1)(x+4)(x-6))/(x-3)#, #x!=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 1218 views around the world You can reuse this answer Creative Commons License