# Formulae to Solve of Polynomial Equations

## The Linear Formula

$$x = {-b \over a}$$

The linear formula gives the solution of $ax+b=0$ for real numbers $a$, $b$ with $a\neq0$.

$$x = {-b\pm\sqrt{b^2-4ac} \over 2a}$$

The quadratic formula gives the solutions of $ax^2+bx+c=0$ for real numbers $a$, $b$, $c$ with $a\neq0$.

## The Cubic Formula

$$x = {-2b + \bigl({-1+\sqrt{-3}\over2}\bigr)^n\root3\of{4\bigl(-2b^3+9abc-27a^2d+\sqrt{(-2b^3+9abc-27a^2d)^2-4(b^2-3ac)^3}\bigr)}+\bigl({-1-\sqrt{-3}\over2}\bigr)^n\root3\of{4\bigl(-2b^3+9abc-27a^2d-\sqrt{(-2b^3+9abc-27a^2d)^2-4(b^2-3ac)^3}\bigr)} \over 6a}$$

The cubic formula gives the solutions of $ax^3+bx^2+cx+d=0$ for real numbers $a$, $b$, $c$, $d$ with $a\neq0$.

Directions: Take $n=0$, $1$, $2$. Use real cube roots if possible, and principal roots otherwise.

## The Quartic Formula

$$\def\sgn{\mathop{\rm sgn}} x = {-3b\pm\biggl(\sqrt{3\Bigl(3b^2-8ac+2a\root3\of{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}+2a\root3\of{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace-\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}\Bigr)}\pm\sqrt{3\Bigl(3b^2-8ac+2a\bigl({-1+\sqrt{-3}\over2}\bigr)\root3\of{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}+2a\bigl({-1-\sqrt{-3}\over2}\bigr)\root3\of{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace-\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}\Bigr)}\biggr)\pm\sgn\biggl(\Bigl(\sgn\bigl(-b^3+4abc-8a^2d\bigr)-{1\over2}\Bigr)\Bigl(\sgn\bigl(\max((2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3,\min(3b^2-8ac,3b^4+16a^2c^2+16a^2bd-16ab^2c-64a^3e))\bigr)-{1\over2}\Bigr)\biggr)\sqrt{3\Bigl(3b^2-8ac+2a\bigl({-1-\sqrt{-3}\over2}\bigr)\root3\of{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}+2a\bigl({-1+\sqrt{-3}\over2}\bigr)\root3\of{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace-\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}\Bigr)}\over12a}$$

The quartic formula gives the solutions of $ax^4+bx^3+cx^2+dx+e=0$ for real numbers $a$, $b$, $c$, $d$, $e$ with $a\neq0$.

Directions: Choose all possibilities for the three $\pm$ signs with the last two equivalent. Use real cube roots if possible, and principal roots otherwise.