Finding the roots of $x^3 - 93x - 308$ extremely quickly? 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!
 A: Vieta's formulas would be the easiest route. Assume the roots are $r_1,r_2,r_3$. The formulas say that $r_1r_2r_3 = 308$, $r_1r_2+r_1r_3+r_2r_3=-93$ and $r_1+r_2+r_3=0$. Since the product of roots is positive and the roots sum to $0$, we must have two negative roots and one positive root. From there it is just a little quick calculation and/or a little luck. In factoring $308$ to make this answer, I choose to do a tree and got $308 = 4*77 = 4*7*11$. That right there gives me the needed numbers since $4+7=11$. So after a double check with synthetic division with 11, I get the answers of $-4,-7,11$.
A: 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).
A: Check out the rational root theorem: http://en.wikipedia.org/wiki/Rational_root_theorem
edit: let $p\over q$ be a root, then $p|a_0$ and $q|a_n$ for $a_nx^n+...+a_0=0$ since $a_n=1$ just check the divisors of 308=2*7*2*11.
A: They probably looked for integer solutions. 
If an integer $n$ solves $n^3-93n-308=0$ then $n(n^2-93)=308$ hence $n$ divides $308=2\times2\times7\times11$. On the other hand, if $|n|\geqslant20$ then $n^2-93\geqslant300$ hence $|n|(n^2-93)\geqslant300|n|\geqslant300\times20$, in particular $n(n^2-93)\ne308$. One is left with the factors of $308$ whose absolute value is less than $20$. 
Furthermore, no integer congruent to $2$ modulo $4$ can solve this because if $n=2\pmod{4}$ then $n^2=0\pmod{4}$ hence $n^2-93=1\pmod{4}$ and $n(n^2-93)=2\pmod{4}$ while $308=0\pmod{4}$. Thus, the only possible roots are $\pm4$, $\pm7$ and $\pm11$.
One can still refine this, noting that $n$ and $n^2-93$ must have the same sign. Since $4^2\lt93$, $7^2\lt93$ and $11^2\gt93$, this reduces the list of candidates to $-4$, $-7$ and $+11$.
Of course, none of this guarantees that these are solutions but the polynomial having integer coefficients with leading coefficient $1$, one knows a priori that every rational solution is actually an integer. The derivative $3n^2-93=3(n^2-31)$ having no rational root, there cannot be any double rational root hence either $-4$, $-7$ and $+11$ would be the three solutions or at least two roots would be irrational. And checking that $-4$, $-7$ and $+11$ are indeed solutions is quite doable in one's head. For example, $n=+11$ gives $n^2-93=121-93=28=4\times7$, bingo.
A: Given that the answered the question so quickly, they must have noticed, as you did, that if the roots are $a,b,c$, then we have $|abc|=308$. Since the last two digits of this are divisible by $4$, so is $308$ divisible by $4$. Very quickly one can figure out that $308=77\cdot 4$ which quickly gives $308=7\cdot 11\cdot 4$. 
There are $3$ roots and because we have $-308$, only one is negative or two of them are negative. We need $a+b+c=0$. Since $7+4=11$, it should come quickly that $-7+-4+11=-11+1=0$ so $-4,-7,11$ are the roots you want.
If you're well adjusted to using the Rational Roots Theorem and are good at mental Math, it's not a leap that one could do this within a few seconds (I'm assuming in the heat of competition time seemed faster than it actually was).
A: Integrability and signs of roots apart (discussed above), the fastness must come from the observation that the product of the three integer roots, $p,q,r$, is $$p\cdot q\cdot r = 308 = 2^2\cdot 7 \cdot 11$$ and that those integer roots must be pairwise relatively coprime as $$p+q+r = 0,$$
which narrows it down to $4, 7,11.$
