How do you factor #(m+n)^2 - p^4#? Algebra Polynomials and Factoring Special Products of Polynomials 1 Answer Konstantinos Michailidis Oct 8, 2015 It is #(m+n)^2-p^4=(m+n+p^2)*(m+n-p^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 # -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 1986 views around the world You can reuse this answer Creative Commons License