I usually factor 3rd degree polynomial in two steps. First, I find all the divisors of the last, coefficient-free part of the polynomial (in this case that's 8) and try (applying Bezout's identity) to plug them into P(x) one by one until I get zero. When I find divisor $d$ that passes the test, I know my polynomial is divisible by $(x - d)$. Then, I do my division and get a quadratic polynomial, which I factor using the formula for solving quadratic equations (I guess I could continue testing via Bezout's but I prefer this way). And that's how I get three factors of 3rd degree polynomial. But in this case, my method fails at step one, because $±1, ±2, ±4, ±8$ do not give zero as a result when plugged into the beginning polynomial.
I also tried factoring it "by hand", meaning, I tried to find common factor between two pairs of the polynomial, but failed. You get $(x−2)x^2−4(x+2)$, so no root there.
How could I solve this? Are candidates for Bezout's test divisors of the free polynomial member only? Can every 3rd degree polynomial be factored in $(x-a)(x-b)(x-c)$ form in R and C?
Thank you in advance.