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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.

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  • $\begingroup$ The roots are very simple, but I have no idea how one would go about finding them. $\endgroup$
    – Arthur
    May 4 '16 at 12:40
  • $\begingroup$ 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. $\endgroup$
    – almagest
    May 4 '16 at 14:02
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Use Horner's method to factorize it : https://en.wikipedia.org/wiki/Horner%27s_method .

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} $, $x= -\frac{1}{7}$, $x=\frac{2}{11}$.

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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

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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 $\frac{2}{11}$.

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This is sort of a remark on Archis Welankar's answer (but is too long to be a comment). The cubic polynomials for which you posted questions seem to be intended to be solved for roots without resorting to heavy computation. The product of the roots is $ \ \mathbf{+}\frac{2}{231} \ $ (there is a slip in that part of the answer I'm citing) and it is a bit suspicious that the denominator has three distinct prime factors, thus $ \ \frac{2}{231} \ = \ \frac{1·1·2}{3·7·11} \ \ . $ So we might entertain the possibility of three distinct rational zeroes. As this product is positive, either all three zeroes are positive, or just one is. The Rule of Signs, however, tells us that there is one positive zero and two negative ones (if they are in fact all real).

Unless the poser of this problem is being particularly cruel, we may go on to consider that these rational zeroes are $ \ \frac{A}{3} \ , \ \frac{B}{7} \ , \ \text{and} \ \frac{C}{11} \ \ , $ with just one of the numerators being positive, two of the numerators being equal to $ \ \pm 1 \ $ and the third equal to $ \ \pm 2 \ . $ We could then use the sum of the zeroes, leaving us to find the appropriate choices of numerators to solve $$ \frac{A}{3} \ + \ \frac{B}{7} \ + \ \frac{C}{11} \ \ = \ -\frac{68}{231} \ \ \Rightarrow \ \ 77A \ + \ 33B \ + \ 21C \ \ = \ -68 \ \ , $$

which only requires a little checking among the available choices of numerators.

Perhaps a bit less work is to instead consider dividing the products of pairs of zeroes by the product of all three zeroes, since $$ \frac{rs \ + \ rt \ + \ st}{rst} \ \ = \ \ \frac{1}{t} \ + \ \frac{1}{s} \ + \ \frac{1}{r} \ \ . $$

Applying this to the putative rational zeroes yields

$$ \frac{-9/231}{2/231} \ \ = \ \ -\frac{9}{2} \ \ = \ \ \frac{3}{A} \ + \ \frac{7}{B} \ + \ \frac{11}{C} \ \ . $$

For the small values of the proposed numerators, this also requires only a bit of testing. By either of these equations, we find $ \ A \ = \ -1 \ \ , \ \ B \ = \ -1 \ \ , \ \ C \ = \ 2 \ \ $ (agreeing with Rebellos's answer).

I make this suggestion to avoid the rather greater amount of effort involved in solving a system of non-linear equations.

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