Cubic Discriminant

We can compute the discriminant of any power of a polynomial. For example, the quadratic discriminant is given by $\Delta_2 = b^2 - 4ac$. But it gets more complicated for higher-degree polynomials.

The discriminant of a cubic polynomial $ax^3 + bx^2 + cx + d$ is given by

$$\Delta_3 = b^2 c^2 - 4ac^3 - 4b^3 d - 27a^2 d^2 + 18abcd.$$

If $\Delta_3 > 0$, then the equation has three distinct real roots.
If $\Delta_3 = 0$, then the equation has a repeated root and all its roots are real.
If $\Delta_3 < 0$, then the equation has one real root and two non-real complex conjugate roots.

Alternatively, we can compute the value of the cubic determinant if we know the roots to the polynomial. Let $\alpha,\beta,$ and $\gamma$ denote the roots of a certain cubic polynomial, then its discriminant is equal to

$$\Delta_3 = a^4(\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2 \; .$$

Here's a good example that requires us to use cubic discriminant:

For a polynomial $P(x)=a_n x^n+a_{n-1} x^{n-1}+\cdots+a_1 x+a_0$ having roots $x_1,x_2,\ldots,x_n$ (counting multiplicity), its discriminant is

$$\Delta=a_n^{2n-2}\prod_{1 \leq i < j \leq n} (x_i-x_j)^2.$$

Notice some things:

• When there's a repeated root, the discriminant vanishes.
• If the coefficients of $P(x)$ are all real, the discriminant is always a real number.
• If all roots are real and distinct, the discriminant is always positive.

For the case when the polynomial is cubic, i.e. $P(x)=ax^3+bx^2+cx+d,$ we get $\Delta=a^4(x_1-x_2)^2(x_2-x_3)^2(x_3-x_1)^2$.

Since the discriminant is symmetric in the roots of the polynomial, we can express it as elementary symmetric polynomials, i.e. the coefficients of $P(x)$. Although there is a general method to derive the discriminant of any polynomial, this is an elementary and algebraic approach.

By Vieta's formula we have

$$\begin{aligned} x_1+x_2+x_3 &=-\dfrac{b}{a}\\\\ x_1x_2+x_1x_3+x_2x_3 &=\dfrac{c}{a}\\\\ x_1x_2x_3&=-\dfrac{d}{a}. \end{aligned}$$

Let's expand the discriminant:

$$\Delta=a^4\Big(\big(x_1x_2^2+x_2x_3^2+x_3x_1^2\big)-\big(x_1^2x_2+x_2^2x_3+x_3^2x_1\big)\Big)^2.$$

Let $m=x_1x_2^2+x_2x_3^2+x_3x_1^2$ and $n=x_1^2x_2+x_2^2x_3+x_3^2x_1$, then $\Delta=a^4(m-n)^2$. We can't directly compute $m$ and $n$, but since they are cyclic polynomials, it turns out that the elementary symmetric polynomials of $m$ and $n$ are symmetric in $x_1,x_2,$ and $x_3$ and hence expressable in terms of the coefficients of $P(x)$. So, let's find them:

$$\begin{aligned} m+n&=x_1x_2(x_1+x_2)+x_2x_3(x_2+x_3)+x_3x_1(x_3+x_1) \\\\ &=(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)-3x_1x_2x_3 \\\\ &=\dfrac{-bc+3ad}{a^2}. \end{aligned}$$

The case for $mn$ requires more tedious expansion:

$$\begin{aligned} mn=&\color{#3D99F6}{(x_1x_2)^3+(x_1x_3)^3+(x_2x_3)^3}+3(x_1x_2x_3)^2+x_1x_2x_3\big(\color{#D61F06}{x_1^3+x_2^3+x_3^3}\color{#3D99F6}\big)\\\\ =&\color{#3D99F6}{(x_1x_2+x_1x_3+x_2x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)(x_1x_2x_3)+3(x_1x_2x_3)^2}+3(x_1x_2x_3)^2\\ &{\color{#3D99F6}{+x_1x_2x_3\big(}}\color{#D61F06}{(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)+3x_1x_2x_3}\color{#3D99F6}\big)\\\\ =&\dfrac{c^3}{a^3}-\dfrac{3bcd}{a^3}+\dfrac{6d^2}{a^2}+\dfrac{b^3d}{a^4}-\dfrac{3bcd}{a^3}+\dfrac{3d^2}{a^2}\\\\ =&\dfrac{ac^3-6abcd+9a^2d^2+b^3d}{a^4}. \end{aligned}$$

Finally, use the identity $(m-n)^2=(m+n)^2-4mn:$

$$\begin{aligned} \Delta &=a^4(m-n)^2=a^4\left(\dfrac{-bc+3ad}{a^2}\right)^2-4a^4\left(\dfrac{ac^3-6abcd+9a^2d^2+b^3d}{a^4}\right)\\\\ &=b^2c^2-6abcd+9a^2d^2-4ac^3+24abcd-36a^2d^2-4b^3d\\\\ &=\boxed{b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd}. \end{aligned}$$

• Cardano's Method

• Quadratic Discriminant

• Root Approximation - Bisection

• Rational Root Theorem