# The uses of the discriminant with cubics

The discriminant for cubic equations is -

Δ​=$$b^2c^2\:−4ac^3\:−4b^3d−27a^2d^2\:+18abcd.$$

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.

• Rational Root Theorem holds for all polynomials. May 29, 2020 at 11:55
• Are the coefficients rational ?
– user65203
May 29, 2020 at 11:56
• Yes the coefficients are rational May 29, 2020 at 12:09
• I know rational root theorem holds for all polynomials, but I was wondering about the role of the discriminant. May 29, 2020 at 12:09
• 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. May 29, 2020 at 12:49

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.