How do you factor # -8x^2 +32#? Algebra Polynomials and Factoring Special Products of Polynomials 1 Answer James May 15, 2018 #-8x^2+32=8[4-x^2]=-8[x^2-4]=8[(x-2)*(x+2)]# Explanation: show below #-8x^2+32=8[4-x^2]=-8[x^2-4]=8[(x-2)*(x+2)]# 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 #x^3-8y^3#? How do you factor # x^3 - 1#? How do you factor #x^3 + y^3#? See all questions in Special Products of Polynomials Impact of this question 5112 views around the world You can reuse this answer Creative Commons License