How do you divide #(2k^3-13k^2-77k+60)div(k-10)# using synthetic division? Precalculus Real Zeros of Polynomials Synthetic Division 1 Answer Cem Sentin Mar 6, 2018 Quotient is #2k^2+7k-7# and remainder is #-10# Explanation: #2k^3-13k^2-77k+60# =#2k^3-20k^2+7k^2-70k-7k+70-10# =#2k^2*(k-10)+7k*(k-10)-7*(k-10)-10# =#(2k^2+7k-7)*(k-10)-10# Hence quotient is #2k^2+7k-7# and remainder is #-10# Answer link Related questions What is synthetic division? What are common mistakes students make with synthetic division? How do I find the quotient and remainder using synthetic division? How do you write the remainder in synthetic division? How do I find the quotient #(x^3+5x^2+x-15)/(x+3)# by using synthetic division? How do I find the roots of a polynomial function by using synthetic division? How can synthetic division be used to factor a polynomial? How do I use synthetic division to find #p(-3)# for #p(x)=x^4-2x^3-4x+4#? Use synthetic division to find #p(4)# for #p(x)=x^4-2x^3-4x+4#? How do you use synthetic division to evaluate #f(3)# given that #f(x)=x^3+2x^2-7x+8#? See all questions in Synthetic Division Impact of this question 1960 views around the world You can reuse this answer Creative Commons License