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

1 Answer
Jan 10, 2018

Answer:

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

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

Explanation:

Given:

#f(x) = x^3-2x^2+3#

Note that:

#f(-1) = -1-2+3 = 0#

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

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

The remaining quadratic has negative discriminant, so can only be factored further using complex coefficients:

#4(x^2-3x+3) = 4x^2-12x+12#

#color(white)(4(x^2-3x+3)) = (2x)^2-2(2x)(3)+3^2+3#

#color(white)(4(x^2-3x+3)) = (2x-3)^2+(sqrt(3))^2#

#color(white)(4(x^2-3x+3)) = (2x-3)^2-(sqrt(3)i)^2#

#color(white)(4(x^2-3x+3)) = ((2x-3)-sqrt(3)i)((2x-3)+sqrt(3)i)#

#color(white)(4(x^2-3x+3)) = (2x-3-sqrt(3)i)(2x-3+sqrt(3)i)#

So:

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