Roots of a Non-Monic Cubic Polynomial

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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.