The discriminant for cubic equations is -


And I am aware that you can determine the number of roots a cubic has using method shown below -

Δ​>0 the equation has three distinct real roots

Δ​=0 the equation has a repeated root and all its roots are real

Δ​<0 the equation has one real root and two non-real complex conjugate roots

But I was wondering if one could determine whether a cubic has rational roots, as you can do with the discriminant for quadratics, and if so what the method would be.

I have noticed that with the cubics I have checked: if the discriminant is a perfect square there are 3 integer solutions, although I have not checked many and I am not sure of the reasoning behind it.

Any help would be greatly appreciated.

  • 1
    $\begingroup$ Rational Root Theorem holds for all polynomials. $\endgroup$ May 29, 2020 at 11:55
  • $\begingroup$ Are the coefficients rational ? $\endgroup$
    – user65203
    May 29, 2020 at 11:56
  • $\begingroup$ Yes the coefficients are rational $\endgroup$
    – user578923
    May 29, 2020 at 12:09
  • $\begingroup$ I know rational root theorem holds for all polynomials, but I was wondering about the role of the discriminant. $\endgroup$
    – user578923
    May 29, 2020 at 12:09
  • $\begingroup$ What would the discriminant help you, say, for $x^3+x+1=0$? It just has no rational root. Here the rational root theorem is much better. $\endgroup$ May 29, 2020 at 12:49

1 Answer 1


The discriminant does contain substantial information about the roots of the cubic. For instance, if $ K/ \mathbf Q $ is the splitting field of your cubic polynomial (the field its roots generate over the rational numbers), then the unique quadratic subfield of this will be $ \mathbf Q(\sqrt{\Delta}) $ where $ \Delta $ is the discriminant of your cubic polynomial. Your observation is therefore correct: if a cubic polynomial has all rational roots, clearly its splitting field can't have a quadratic subfield, so $ \Delta $ must be a perfect square. A more concrete way to see this is in terms of the expression

$$ \Delta = \prod_{1 \leq i < j \leq n} (\alpha_j - \alpha_i)^2 $$

for the discriminant of a degree $ n $ polynomial $ P(x) $ whose roots are $ \alpha_1, \alpha_2, \ldots \alpha_n $. If the $ \alpha_i $ are all rational, $ \Delta $ is clearly a perfect square in $ \mathbf Q $, and the same is true over the integers.

However, you can't determine if a cubic has rational roots or not by looking at the discriminant alone. For instance, the polynomial $ P(x) = x^3 + x^2 - 2x - 1 $ has discriminant $ 49 = 7^2 $ and yet it has no rational roots. The discriminant being a perfect square tells you that the Galois group of the cubic polynomial is a subgroup of $ A_3 $, so it must either have three rational roots or none, but you can't conclude anything further in general.


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