How do you simplify #(2c^2+2c-12)/(-8+2c+c^2)#? Algebra Rational Equations and Functions Division of Polynomials 1 Answer EZ as pi Oct 4, 2016 #(2(c+3))/(c+4)# Explanation: #(2c^2+2c-12)/(-8+2c+c^2)# =#(2(c^2+c-6))/(c^2 +2c-8)" " (larr"common factor of 2")/(larr"re-arrange the terms")# #=(2(c+3)cancel((c-2)))/((c+4)cancel((c-2))) " "larr# find factors of quadratics #=(2(c+3))/(c+4)# 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 1347 views around the world You can reuse this answer Creative Commons License