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. $\endgroup$ Commented May 29, 2020 at 11:55
  • $\begingroup$ Are the coefficients rational ? $\endgroup$
    – user65203
    Commented May 29, 2020 at 11:56
  • $\begingroup$ Yes the coefficients are rational $\endgroup$
    – user578923
    Commented 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
    Commented 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. $\endgroup$ Commented 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .