ionI always start by checking if 1 or -1 is a solution by substituting them into the equation and seeing if The result is zero ( Neither of them is ). Then I check the other rational solutions which are +/- 3 and +/-9 using synthetic division. -3 is a solution[so (x+3) is a factor] with a remainder of x^3 +3x^2 -+x +3 which can be factored by grouping to (x^2 +1)(x+3)
‘That gives us (x+3)(x+3)(x^2 + 1) (X^2 + 1) can be factored to (x+i)(x-i)
Final answer is (x+3)(x+3)(x+i)(x-i)