How do you factor #4f^2 - 64#? Algebra Polynomials and Factoring Special Products of Polynomials 1 Answer Himanshu Shekhar Jun 12, 2016 # 4(f^2-16) # Explanation: Take out 4 as common from the expression. # 4f^2 -64 # # 4(f^2-16) # Answer link Related questions What are the Special Products of Polynomials? What is a perfect square binomial and how do you find the product? How do you simplify by multiplying #(x+10)^2#? How do you use the special product for squaring binomials to multiply #(1/4t+2 )^2#? How do you use the special product of a sum and difference to multiply #(3x^2+2)(3x^2-2)#? How do you evaluate #56^2# using special products? How do you multiply #(3x-2y)^2#? How do you factor # -8x^2 +32#? How do you factor #x^3-8y^3#? How do you factor # x^3 - 1#? See all questions in Special Products of Polynomials Impact of this question 1790 views around the world You can reuse this answer Creative Commons License