TITLE: Finding the roots of $x^3 - 93x - 308$ extremely quickly?
QUESTION [10 upvotes]: I was at a quiz bowl competition and one of the questions was to find the roots of this polynomial. In one or two seconds after reading the question, somebody on the other team buzzed and got the right answer $(-4, -7, 11)$. How is this possible? What did he use to get the answer? This is a high school competition.
One way I suppose he could've done it is to realize that, since there is no $x^2$ term, the sum of the roots is 0, then find three integers $a, b, c$ such that $abc = 308$, but doing all of that in  1 or 2 seconds? is that possible?
Okay, so from the comments I see that it was just some really fast mental math, not some kind of advanced formula that I'm not aware of. Thanks for the help!

REPLY [3 votes]: The same reasoning yields the following general result. The slight abstraction has the advantage of making it much clearer exactly why the argument works.
Theorem $\ $ If $\ f(x) = x^3+ dx - abc\in\Bbb Z[x]\,$ has $3$ integer roots, and $\,a<b <c,\,$ are distinct primes $ $ (or squares of such), $ $ and  if $\,f(\pm1)\ne 0,\, $ then the roots of $\,f(x)\,$ are $\,-a,\, -b,\,\ c.$
Proof $\ $ By the Rational Root Test and Vieta the roots are integers with product $\,abc.\,$ By Vieta, the sum of the roots $= 0\,$ so the roots are pairwise coprime (else prime $p\mid r_1,r_2\,\Rightarrow\,p\mid r_3 = -r_1-r_2$ so $\,p^3\mid r_1r_2r_3= abc,\,$ contra hypotheses).  Since $\,a,b,c\,$ are powers of distinct primes, by unique factorization, the only way $\,abc\,$ can be factored into $\,3\,$ pairwise coprime factors $(\ne \pm1,\,$ since $f(\pm1)\ne 0)$ is if those factors are $\,a,b,c\,$ up to signs. 
Since the roots have sum $= 0$ and product  $>0$ there are two negative roots and one positive root (necessarily the root $\,c\,$ of largest magnitude).