How do you factor #2w^3 + 54#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer mason m Dec 19, 2015 #2(w+3)(w^2-3w+9)# Explanation: First, factor out a common #2#. #=2(w^3+27)# Notice that #(w^3+27)# is a sum of cubes, which follows the rule #a^3+b^3=(a+b)(a^2-ab+b^2)# Thus, #2(w^3+27)=2((w)^3+(3)^3)=2(w+3)(w^2-3w+9)# 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 1633 views around the world You can reuse this answer Creative Commons License