# Factoring polynomial $x^3−2x^2−4x−8$ that fails Bezout's identity test

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?

• $x^3+x$ can't be factored over the reals... – JP McCarthy Jan 28 '15 at 14:20
• One thing is sure with third degree polynomials: they have at least one real root. But take $x^3+1$... How many real roots? – Martigan Jan 28 '15 at 14:21
Your polynomial is irreducible over $\mathbb Q[x]$. Its only real root is irrational, namely $$\frac{2+\sqrt[3]{152-24\sqrt{33}}+2\sqrt[3]{19+3\sqrt{33}}}3\not\in\mathbb Q$$ The same goes for the other two, whose expressions can also be deduced using the cubic formula. Perhaps you are confused by thinking that cubic conjugate roots come in pairs, just like quadratic conjugate roots. They don't. They come in triples.