How do you factor #x^3-3x^2-4x+12#? Algebra Polynomials and Factoring Factoring by Grouping 1 Answer EZ as pi May 3, 2017 Answer: #=(x-3)(x+2)(x-2)# Explanation: #x^3 -3x^2 -4x +12" "larr# group into pairs #=(x^3 -3x^2)+ (-4x +12)" "larr# there must be a #+# between. #x=^2(x-3)color(blue)(-4(x-3))" "larr div color(blue)(-4)# out as the factor to change the signs. #=(x-3)(x^2-4)" "larr# factors further as difference of squares. #=(x-3)(x+2)(x-2)# Related questions What is Factoring by Grouping? How do you factor by grouping four-term polynomials and trinomials? Why does factoring polynomials by grouping work? How do you factor #2x+2y+ax+ay#? How do you factor #3x^2+8x+4# by using the grouping method? How do you factor #6x^2-9x+10x-15#? How do you group and factor #4jk-8j^2+5k-10j#? What are the factors of #2m^3+3m^2+4m+6#? How do you factor quadratics by using the grouping method? How do you factor #x^4-2x^3+5x-10#? See all questions in Factoring by Grouping Impact of this question 167 views around the world You can reuse this answer Creative Commons License