I'm trying to solve this equation:

$x^3 + 3x^2 + x + 1 = 0$

I tried to guess a root so I could try polinomial division to try factorise this equation but I couldn't find any whole integers for that. I thought of maybe using variable substitution but I dont think that would work here.

Any tips/explained solution on how to solve an equation like this? Thanks!

  • $\begingroup$ You can use the formula for the roots of a cubic equation. en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots $\endgroup$ – Ángel Mario Gallegos May 20 '16 at 17:54
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    $\begingroup$ It has one real root near -2.7693 and two complex roots. You just have to use the standard formulae. $\endgroup$ – almagest May 20 '16 at 17:55
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    $\begingroup$ The $3$ is deliberate, as an integer change makes a "depressed" cubic, no zero term. Take $x = t-1,$ you get $t^3 + pt + q$ for integers $p,q,$ no $t^2$ term. $\endgroup$ – Will Jagy May 20 '16 at 18:11

Wolfy says one real, two complex roots, so you are out of luck if you want a simple solution.

The real root is, after some reformatting,

$ -1-\dfrac{(9-\sqrt{57})^{1/3}}{3^{2/3}}-\dfrac{2}{(3 (9-\sqrt{57}))^{1/3}} $


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