Question

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

Answer 1

`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)`