How do you factor #49x^2 - 121y^2#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Don't Memorise May 4, 2015 #49x^2 - 121y^2# can be written as #(7x)^2 - (11y)^2# this is of the form: #a^2 - b^2 = ( a + b) (a -b ) # here: #a = 7x# and # b = 11y# so , # (7x)^2 - (11y)^2= (7x+11y) (7x- 11y) # the factorized form of #49x^2 - 121y^2# is: #(7x+11y) (7x- 11y)# Answer link Related questions How do you factor special products of polynomials? How do you identify special products when factoring? How do you factor #x^3 -8#? What are the factors of #x^3y^6 – 64#? How do you know if #x^2 + 10x + 25# is a perfect square? How do you write #16x^2 – 48x + 36# as a perfect square trinomial? What is the difference of two squares method of factoring? How do you factor #16x^2-36# using the difference of squares? How do you factor #2x^4y^2-32#? How do you factor #x^2 - 27#? See all questions in Factor Polynomials Using Special Products Impact of this question 4675 views around the world You can reuse this answer Creative Commons License