How do you divide #( 6x^2 - 4x - 24 )/(2x+2)#? Algebra Rational Equations and Functions Division of Polynomials 1 Answer SagarStudy Dec 28, 2015 #(6x^2-4x-24)/(2x+2)=(2(3x^2-2x-12))/(2(x+1))# #=(3x^2-2x-12)/(x+1)# 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 1146 views around the world You can reuse this answer Creative Commons License