How do you simplify # [ (x^2-1)^3 (2x+1) ]#? Algebra Polynomials and Factoring Multiplication of Polynomials by Binomials 1 Answer Mark D. Jun 25, 2018 #(x^2-1)^3=(x^2-1)(x^2-1)(x^2-1)# #(x^2-1)(x^2-1)=x^4-2x^2+1# #(x^4-2x^2+1)(x^2-1)=x^6-2x^4+x^2-x^4+2x^2-1# #=x^6-3x^4+3x^2-1# #[(x^2-1)^3(2x+1)]=(x^6-3x^4+3x^2-1)(2x+1)# =#2x^7-6x^5+6x^3-2x+x^6-3x^4+3x^2-1# =#2x^7+x^6-6x^5-3x^4+6x^3+3x^2-2x-1# Answer link Related questions What is FOIL? How do you use the distributive property when you multiply polynomials? How do you multiply #(x-2)(x+3)#? How do you simplify #(-4xy)(2x^4 yz^3 -y^4 z^9)#? How do you multiply #(3m+1)(m-4)(m+5)#? How do you find the volume of a prism if the width is x, height is #2x-1# and the length if #3x+4#? How do you multiply #(a^2+2)(3a^2-4)#? How do you simplify #(x – 8)(x + 5)#? How do you simplify #(p-1)^2#? How do you simplify #(3x+2y)^2#? See all questions in Multiplication of Polynomials by Binomials Impact of this question 1234 views around the world You can reuse this answer Creative Commons License