Question #3828f Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Cesareo R. Apr 17, 2017 See below. Explanation: #a x^2+bx+c = a(x^2+b/ax+c/a)# and now #x =1/2 (-b/apm sqrt((b/a)^2-4(c/a)))# or #ax^2+bx+c=a(x+b/(2a)+1/2sqrt((b/a)^2-4(c/a)))(x+b/(2a)-1/2sqrt((b/a)^2-4(c/a)))# or #ax^2+bx+c=a(x+b/(2a)+1/(2absa)sqrt(b^2-4ac))(x+b/(2a)-1/(2absa)sqrt(b^2-4ac))# 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 1322 views around the world You can reuse this answer Creative Commons License