How do you divide #(x^2-1/6x-1/6)/(x-1/2)#? Precalculus Real Zeros of Polynomials Long Division of Polynomials 1 Answer Shwetank Mauria Apr 16, 2016 #(x^2-1/6x-1/6)/(x-1/2)=x+1/3# Explanation: #(x^2-1/6x-1/6)/(x-1/2)=(1/6(6x^2-x-1))/(1/2(2x-1))=(2(6x^2-x-1))/(6(2x-1))# = #(6x^2-x-1)/(3(2x-1)# Now splitting the middle term #-x# in numerator to #-3x+2x#, we get = #(6x^2-3x+2x-1)/(3(2x-1))# = #(3x(2x-1)+1(2x-1))/(3(2x-1))# = #((3x+1)(2x-1))/(3(2x-1))=(3x+1)/3=x+1/3# Answer link Related questions What is long division of polynomials? How do I find a quotient using long division of polynomials? What are some examples of long division with polynomials? How do I divide polynomials by using long division? How do I use long division to simplify #(2x^3+4x^2-5)/(x+3)#? How do I use long division to simplify #(x^3-4x^2+2x+5)/(x-2)#? How do I use long division to simplify #(2x^3-4x+7x^2+7)/(x^2+2x-1)#? How do I use long division to simplify #(4x^3-2x^2-3)/(2x^2-1)#? How do I use long division to simplify #(3x^3+4x+11)/(x^2-3x+2)#? How do I use long division to simplify #(12x^3-11x^2+9x+18)/(4x+3)#? See all questions in Long Division of Polynomials Impact of this question 1432 views around the world You can reuse this answer Creative Commons License