Question #98f7b Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer smendyka Dec 12, 2016 #x^2 + (9/16)x^2# transforms into #(25/16)x^2# as shown below. Explanation: To factor this expression gives: #(1 + 9/16)x^2# To add 1 and #9/16# we must get this over a common denominator of #16#: #16/16*1 + 9/16)x^2 -># #(16/16 + 9/16)x^2 -># #(25/16)x^2# 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 1226 views around the world You can reuse this answer Creative Commons License