How do you multiply #(3y-8)^2#? Algebra Polynomials and Factoring Multiplication of Polynomials by Binomials 1 Answer Gazza Mar 23, 2018 #=9y^2-48y+64# Explanation: #(3y - 8)^2 #=(3y - 8)(3y - 8)# Use the distributive property #=(3y)(3y)+(3y)(-8)+(-8)(3+(-8)(-8)# #=9y^2-24y-24y +64# #=9y^2-48y+64# 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 2157 views around the world You can reuse this answer Creative Commons License