# Roots of a Non-Monic Cubic Polynomial

Find all roots of $f(x)=231x^3+68x^2-9x-2$

I cannot use the cubic formula or Viete's theorem here because the polynomial is not monic. The only other way I can think of doing this is by the rational roots theorem. My only concern is that there are many divisors of 231, namely 231, -231, 1, -1, 3, 3, 77, -77, 11, -11, 7, -7.

Other than the rational roots theorem, is there any way to solve this? There might not be any easy methods to do so, but I am just wondering if I can save some time.

Thanks.

• The roots are very simple, but I have no idea how one would go about finding them. – Arthur May 4 '16 at 12:40
• If you suspect that the roots are simple fractions then you need the numerator to divide 2 and the denominator to divide $231=3\cdot7\cdot11$. In this case $x=\pm1,\pm2$ are obviously not roots, so you are looking for factors like $(11x\pm2),(11x\pm1),(7x\pm1)$ etc. They are not hard to find. – almagest May 4 '16 at 14:02

The factorized polynomial is : $f(x) = (3 x+1) (7 x+1) (11 x-2)$ and the solutions of course are : $x = -\frac{1}{3}$, $x= -\frac{1}{7}$, $x=\frac{2}{11}$.
Hint use general theory of equation. Sum of roots is $-68/231$, product of roots $-2/231$ and $ab+bc+ac=-9/231$ and solving three simultaneous equations
You can use Hensel lifting to find solutions modulo powers of 3. E.g. using the solution $x = 1$ modulo 3, Hensel lifting yields that modulo 9 there is a solution for $x = 1$, and it becomes $x = -44$ modulo 81 and it becomes $x = 2386$ modulo $81^2$. You can then use the rational reconstruction method to find the fraction $\frac{r}{s}$ such that $r s^{-1} \bmod 81^2$ yields $2386$. This will work provided twice the product of the bounds on the numerator and denominator is smaller than $81^2$. Since $4\times 231 < 81^2$ this will work.
What we then need to do is run the extended Euclidean algorithm, this yields intermediary results that can be written as equations of the form $A 2386 + B 81^2 = C$, which we can interpret as $2386 = C A^{-1} \bmod 81^2$. You run the algorithm until you get small numbers for both $C$ and $A$ (there is always going to be a pair such that twice the product of the numbers falls below the modulo that one is using). In this case one finds that $2386\bmod 81^2$ corresponds to $\frac{2}{11}$.