That doesn't factor neatly. Applying the quadratic equation, we find two imaginary solutions: Zero plus or minus i times the square root of 3.
x = 1.7320508075688772i
x = -1.7320508075688772i
where i is the square root of negative one.
x3 + x2 - 3x - 3 x(x2 + x - 3) - 3
3 - 3x + x2 - x3 = (1 - x)(x2 + 3)
(x + 3)(x + 3)
x2 + 8x + 15 = (x + 3)(x + 5)
x2 + 8x + 15 = (x + 3) (x + 5).
3(x2 + 3x + 1)
(x+3)(x+3)
x2-6x+9 = (x-3)(x-3) when factorised.
(x - 1)(x + 3)
(x-1)(x-3)
3(x2 - 2x + 3)
(x - 3)(x - 3)