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

1 Answer
May 13, 2016

Answer:

#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: #-1-1-1+3 = 0#

So #x=-1# is a zero and #(x+1)# a factor:

#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)#