How do you factor #36x^2-49#? Algebra Polynomials and Factoring Special Products of Polynomials 1 Answer Meave60 Jun 28, 2015 Use the difference of squares equation: #(a^2-b^2)=(a+b)(a-b)#. Identify #a# and #b#, and substitute the values into the equation. Explanation: #36x^2-49# is in the form of the difference of squares: #(a^2-b^2)=(a+b)(a-b)# #a=6x# #b=7# #36x^2-49=(6x+7)(6x+7)# 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 11694 views around the world You can reuse this answer Creative Commons License