How do you factor completely #12y^3 + 33y^2 - 6y #?
1 Answer
Aug 28, 2016
Explanation:
#12y^3+33y^2-6y#
#=3y(4y^2+11y-2)#
#=3y((2y)^2+2(2y)(11/4)+(11/4)^2-(11/4)^2-2)#
#=3y((2y+11/4)^2-153/4^2)#
#=3y((2y+11/4)^2-(sqrt(153)/4)^2)#
#=3y(2y+11/4-sqrt(153)/4)(2y+11/4+sqrt(153)/4)#
