\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\begin{document}
\title{Solving a Quartic Equation With No Rational Roots}
\author{Amit Yaron}
\maketitle
I once saw a nice quartic equation on YouTube:
\begin{equation*}
x^4-2x^3+x=6
\end{equation*}
But I didn't wat to watck the video. I want to solve it myself using the method
commonly used on those videos: factorizing the quartic polynomial into
2 quadratic ones. The factorizarion is:
\begin{equation}
  (x^2-ax+b)[x^2+(a-2)x-\frac6b]=x^4-2x^3+x-6
\end{equation}
hoping to find integer $a$ and $b$.
But, before trying to find them, let us talk about the possible number of
values for pairs of complex $a$ and $b$:
accoring to the fundamental law of algebra a quartic polynomial $P(z)$ can be factored over $\mathbb{C}$ as:
\begin{equation*}
  P(z)=a(z-z_1)(z-z_2)(z-z_3)(z-z_4)
\end{equation*}
In how many ways we can choose order pairs of 2 unordered pairs from the 4 factors?
\begin{equation*}
  \binom 4 2 = 6
\end{equation*}
6 ways, and with the Commutativw Law, we actually have 3. So, in (1):\\
$-a$ can be a root of one cubic equation and $a-2$ of another.
Let us now expand:
\begin{equation*}
  x^4-2x^3+x-6=x^4-2x^3+[b-\frac6b-a(a-2)]x^2+[(a-2)b+\frac{6a}b]x-6
\end{equation*}
By equating the coefficient, we obtain the following system:
\begin{equation}
  \left\{
  \begin{alignedat}{1}
    b-\frac6b-a(a-2)&=0\\(a-2)b+\frac{6a}b&=1
  \end{alignedat}
  \right.
\end{equation}
Let us get rid of the $b$ in the denominator. Multiply the first wquation of (2) by $a$
\begin{equation}
  ab-\frac{6a}b-a^2(a-2)=0
\end{equation}
Add (3) to the second equation of(2):
\begin{align*}
  &ab-a^2(a-2)+(a-2)b=1\Rightarrow\\
  \Rightarrow &ab-a^3+2a^2+ab-2b=1\Rightarrow\\
  \Rightarrow &-a^3+2a^2+2ab-2b=1\Rightarrow\\
  \Rightarrow &a^3-2a^2-2ab+2b+1=0\Rightarrow\\
  \Rightarrow &a^3-1-2a^2-2ab+2b+2=0\Rightarrow\\
  \Rightarrow &(a-1)(a^2+a+1)-2(a^2-1)-(2a-2)b=0\Rightarrow\\
  \Rightarrow &(a-1)(a^2+a+1)-(2a+2)(a-1)-2b(a-1)=0\Rightarrow\\
  \Rightarrow &(a-1)(a^2-a-1-2b)=0
\end{align*}
Luckily, we find that $a=1$ is a solution. Let us find the corresponding values of $b$: from the equations of (2).

Equation 1:
\begin{align*}
  &b-\frac6b-1(1-2)=0\Rightarrow\\
  \Rightarrow &b-\frac6b+1=0\Rightarrow\\
  \Rightarrow &b^2-6+b=0\Rightarrow\\
  \Rightarrow &b^2+b-6=0\Rightarrow\\
  \Rightarrow &(b-2)(b+3)=0
\end{align*}

Equation 2:
\begin{align*}
  &(1-2)b+\frac6b=1\Rightarrow\\
  \Rightarrow &-b+\frac6b=1\Rightarrow\\
  \Rightarrow &b^2-6=-b\Rightarrow\\
  \Rightarrow &b^2+b-6=0\Rightarrow\\
  \Rightarrow &(b-2)(b+3)=0
\end{align*}
\\
Cool!, the solutions are $b\in(2,-3)$,
Plug the values into (1) and get:
\begin{equation}
  (x^2-x+2)(x^2-x-3)=0
\end{equation}

The solutions from $x^2-x+2=0$ are:
\begin{align*}
  &4x^2-4x+8=0\Rightarrow\\
  \Rightarrow &4x^2-4x+1=-7\Rightarrow\\
  \Rightarrow &(2x-1)^2=(\sqrt7i)^2\Rightarrow\\
  \Rightarrow &2x-1=\pm\sqrt7i\Rightarrow\\
  \Rightarrow &x=\frac{1\pm\sqrt7i}2
\end{align*}

The solutions from $x^2-x-3=0$ are:
\begin{align*}
  &4x^2-4x-12=0\Rightarrow\\
  \Rightarrow &4x^2-4x+1=13\Rightarrow\\
  \Rightarrow &(2x-1)^2=(\sqrt{13})^2\Rightarrow\\
  \Rightarrow &2x-1=\pm\sqrt{13}\Rightarrow\\
  \Rightarrow &x=\frac{1\pm\sqrt{13}}2
\end{align*}

\end{document}