The Roots of Polynomials

An electron can be thought of as a ball that's spinning, except it's not a ball and it's not spinning.

This is a collection of programs that gives the roots (i.e., zeros) of polynomials of varying degrees. Here, these are not solved numerically—that's boring. Up to degree four (quartic), this can be done exactly. Now you can calculate solutions whilst waiting in line for coffee.

Linear
Quadratic
Cubic
Quartic
Quintic
Sextic
Septic

If real, use the plots to interactively verify the zeros of each calculation.

Linear

A linear equation in one variable $x$ can be written as $a x + b = 0$, with $a \ne 0$. The solution is $x = -b / a$.

$a =$

$b =$

$x =$

Code: Fortran; JavaScript

Quadratic

A quadratic equation in one variable $x$ can be written as $a x^2 + b x + c = 0$, with $a \ne 0$. The solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}$. Completing the square is one of several ways for deriving this formula. To learn why the quadratic formula works and why quadratics are easier to solve than cubics, please see The Symmetry That Makes Solving Math Equations Easy.

$a =$

$b =$

$c =$

$x_1 =$ $i$

$x_2 =$ $i$

Code: Fortran; JavaScript

Cubic

A cubic equation in one variable $x$ can be written as $a x^3 + b x^2 + c x + d = 0$, with $a \ne 0$. The solutions for the general cubic have the form $x_k = -\frac{1}{3 a} \left( b + \xi^k C + \frac{\Delta_0}{\xi^k C} \right)$, $k \in \{0, 1, 2\}$, with $k = 0$ always returning a real solution. A complete description of this method is given here. Note the term $\frac{\Delta_0}{\xi^k C}$ when applying PEMDAS. For those interested in a bit of history surrounding the cubic formula, please see The Sordid Past of the Cubic Formula.

$a =$

$b =$

$c =$

$d =$

$x_1 =$ $i$

$x_2 =$ $i$

$x_3 =$ $i$

Code: Fortran; JavaScript

Quartic

A quartic equation in one variable $x$ can be written as $a x^4 + b x^3 + c x^2 + d x + e = 0$, with $a \ne 0$. The solutions for the general quartic are described in the Notes.

$a =$

$b =$

$c =$

$d =$

$e =$

$x_1 =$ $i$

$x_2 =$ $i$

$x_3 =$ $i$

$x_4 =$ $i$

Code: Fortran; JavaScript

Notes

gfortran -g -std=f2018 -Wall -Wextra -O2 -fall-intrinsics -o <program> <program>.f08

Solutions of the General Quartic

We wish to obtain the four solutions of the general quartic $\alpha x^4 + \beta x^3 + \gamma x^2 + \delta x + \epsilon = 0$. Start by reformulating the general quartic into a depressed quartic $x^4 + p x^2 + q x + r = 0$, with coefficients $p = (8 \gamma \alpha - 3 \beta^2) / 8 \alpha^2$, $q = (\beta^3 - 4 \gamma \beta \alpha + 8 \delta \alpha^2) / 8 \alpha^3$, and $r = (-3 \beta^4 + 256 \epsilon \alpha^3 - 64 \delta \beta \alpha^2 + 16 \gamma \beta^2 \alpha) / 256 \alpha^4$. Next, formulate the depressed quartic into a resolvent cubic $a x^3 + b x^2 + c x + d = 0$, with coefficients $a = 8$, $b = -4 p$, $c = -8 r$, and $d = 4 p r - q^2$. Now find the real root $s_1$ of the resolvent cubic using the general cubic formula. Arbitrarily choosing +$R$ in $C = \sqrt[3]{(\Delta_1 + R) / 2}$, we will have to test radical, $R = \sqrt{\Delta_1^2 - 4 \Delta_0^3}$, for NaN (not a number) if the resolvent cubic has three real solutions. As an aside, JS is somehow not good at determining NaN, and so requires a little hand-holding:

if (parseInt(Cardano[0] / math.conj(Cardano[0]), 10) === 1) {
    if (parseInt(Cardano[1] / math.conj(Cardano[1]), 10) === 1) {
        if (parseInt(Cardano[2] / math.conj(Cardano[2]), 10) === 1) {
            for (j = 0; j <= 2; j = j + 1) {
                if (math.re(Cardano[j]) > 0.0) {
                    s_1 = math.re(Cardano[j]);
                    radical = math.sqrt(2.0 * s_1 - p);
                    break;
                }
            }
        }
    }
}

Note that three real solutions of the resolvent cubic always returns four complex solutions of the general quartic. Finally, the four roots of the general quartic are

$\displaystyle{ x_1 = -\frac{1}{2} \left( R - \sqrt{-2 s_1 - p + \frac{2 q}{R}} \right) - \frac{\beta}{4 \alpha} \text{,} }$

(N.

$\displaystyle{ x_2 = -\frac{1}{2} \left( R + \sqrt{-2 s_1 - p + \frac{2 q}{R}} \right) - \frac{\beta}{4 \alpha} \text{,} }$

(N.

$\displaystyle{ x_3 = \frac{1}{2} \left( R + \sqrt{-2 s_1 - p - \frac{2 q}{R}} \right) - \frac{\beta}{4 \alpha} \text{,} }$ and

(N.

$\displaystyle{ x_4 = \frac{1}{2} \left( R - \sqrt{-2 s_1 - p - \frac{2 q}{R}} \right) - \frac{\beta}{4 \alpha} \text{.} }$

(N.

Solving quintic equations in terms of radicals ($n$th roots, $\sqrt[n]{x}$) was a major problem in algebra from the 16th century until the first half of the 19th century, when the impossibility of such a general solution was proved with the Abel-Ruffini theorem.

Contents

Linear

Quadratic

Cubic

Quartic