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

Question

1345 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-13T10:58:45

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

$= (x+1)(x-1-sqrt(2)i)(x-1+sqrt(2)i)$

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:

$x^2-2x+3$

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

$= (x-1)^2+2$

$= (x-1)^2-(sqrt(2)i)^2$

$= ((x-1)-sqrt(2)i)((x-1)+sqrt(2)i)$

$= (x-1-sqrt(2)i)(x-1+sqrt(2)i)$

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