How do you factor #x^3 - x^2 +x +3#?
1 Answer
May 13, 2016
#x^3-x^2+x+3 = (x+1)(x^2-2x+3)#
#= (x+1)(x-1-sqrt(2)i)(x-1+sqrt(2)i)#
Explanation:
Notice that if the signs of the coefficients on the terms of odd degree are inverted then the sum of the coefficients is zero.
That is:
So
#x^3-x^2+x+3 = (x+1)(x^2-2x+3)#
The discriminant of the remaining quadratic factor is negative, but we can factor it with Complex coefficients by completing the square:
#x^2-2x+3#
#= (x-1)^2-1+3#
#= (x-1)^2+2#
#= (x-1)^2-(sqrt(2)i)^2#
#= ((x-1)-sqrt(2)i)((x-1)+sqrt(2)i)#
#= (x-1-sqrt(2)i)(x-1+sqrt(2)i)#
