How do you divide #(x^2+12x-18)/(x-3)#? Algebra Rational Equations and Functions Division of Polynomials 1 Answer Daniel · Gazza Mar 8, 2018 #x + 9# Explanation: #x-3# divided into #x^2+12x-18# First part of answer is #x# #x-3# times #x# is #x^2-3x# Subtract from #x^2+12x-18# to get #9x-18# Second part will then be #9# #x-3# times #9# is #9x-18# So answer is #x+9# 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 1811 views around the world You can reuse this answer Creative Commons License