How do you factor completely: #21x^3 + 35x^2 + 9x + 15 #? Algebra Polynomials and Factoring Factoring Completely 1 Answer George C. · Antoine Jul 29, 2015 Factor by grouping to find: #21x^3+35x^2+9x+15=(7x^2+3)(3x+5)# Explanation: #21x^3+35x^2+9x+15# #=(21x^3+35x^2)+(9x+15)# #=7x^2(3x+5)+3(3x+5)# #=(7x^2+3)(3x+5)# The term #7x^2+3# has no linear factors with real coefficients since #7x^2+3 >= 3 > 0# for all #x in RR#. 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 813 views around the world You can reuse this answer Creative Commons License