How do you factor completely #3x^3y^2 - 3x^2y^2 + 3xy^2#?

1 Answer
Apr 7, 2016

Answer:

#3x^3y^2-3x^2y^2+3xy^2#

#= 3xy^2(x^2-x+1)#

#= 3xy^2(x-1/2-sqrt(3)/2i)(x-1/2+sqrt(3)/2i)#

Explanation:

All of the terms are divisible by #3xy^2#, so separate that out as a factor first:

#3x^3y^2-3x^2y^2+3xy^2 = 3xy^2(x^2-x+1)#

The remaining quadratic factor is of the form #ax^2+bx+c# with #a=1#, #b=-1# and #c=1#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-1)^2 - (4*1*1) = 1-4 = -3#

Since this is negative the quadratic has no factors with Real coefficients.

If we allow Complex coefficients then we can factor it further.

#x^2-x+1 = 0# when

#x = (-b+-sqrt(b^2-4ac))/(2a)#
#=(-b+-sqrt(Delta))/(2a)#

#=(1+-sqrt(-3))/2#

#=1/2+-sqrt(3)/2i#

Hence we find:

#x^2-x+1 = (x-1/2-sqrt(3)/2i)(x-1/2+sqrt(3)/2i)#