How to solve this?#f=(2x^2-x-1)^10#Calculate #x_1+x_2+x_3+...+x_20#. Precalculus Polynomial Functions of Higher Degree Zeros 1 Answer Cesareo R. Mar 26, 2017 #5# Explanation: #2x^2-x-1 =2 (x+1/2)(x-1)# and #(2x^2-x-1)^10=2^10(x+1/2)^10(x-1)^10# so the sum of their #20# roots is #x_1+x_2+cdots+x_20 = 10 xx 1-10 xx 1/2 = 5# Answer link Related questions What is a zero of a function? How do I find the real zeros of a function? How do I find the real zeros of a function on a calculator? What do the zeros of a function represent? What are the zeros of #f(x) = 5x^7 − x + 216#? What are the zeros of #f(x)= −4x^5 + 3#? How many times does #f(x)= 6x^11 - 3x^5 + 2# intersect the x-axis? What are the real zeros of #f(x) = 3x^6 + 1#? How do you find the roots for #4x^4-26x^3+50x^2-52x+84=0#? What are the intercepts for the graphs of the equation #y=(x^2-49)/(7x^4)#? See all questions in Zeros Impact of this question 1401 views around the world You can reuse this answer Creative Commons License