How do you factor completely: #2x^2 - 6x - 56#? Algebra Polynomials and Factoring Factoring Completely 1 Answer George C. Jul 18, 2015 Separate out the scalar factor #2# then find pair of numbers #4# and #7# whose product is #28# and difference #3#, hence... #2x^2-6x-56 = 2(x-7)(x+4)# Explanation: #2x^2-6x-56 = 2(x^2-3x-28)# To factor #x^2-3x-28#, find a pair of numbers whose product is #28# and whose difference is #3#. The pair #7#, #4# works. So #x^2-3x-28 = (x-7)(x+4)# and #2x^2-6x-56 = 2(x-7)(x+4)# Answer link Related questions What is Factoring Completely? How do you know when you have completely factored a polynomial? Which methods of factoring do you use to factor completely? How do you factor completely #2x^2-8#? Which method do you use to factor #3x(x-1)+4(x-1) #? What are the factors of #12x^3+12x^2+3x#? How do you find the two numbers by using the factoring method, if one number is seven more than... How do you factor #12c^2-75# completely? How do you factor #x^6-26x^3-27#? How do you factor #100x^2+180x+81#? See all questions in Factoring Completely Impact of this question 4323 views around the world You can reuse this answer Creative Commons License