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\begin{document}
\title{Analytic Methods of Finding Roots of Polynomial Equations}
\author{Kevin Iga}
\date{August 10, 1992}
\maketitle

\section{Introduction}
This document summarizes some of the more useful methods of solving
integer, rational, real, and complex polynomial equations using analytic
means.

\section{Quadratic}

The solution to the quadratic equation in one variable depends on a method
called ``completing the square''.  The idea of completing the square is
to translate the independent variable so that the linear term goes to zero.

The general quadratic equation in one variable is:

\[ax^2+bx+c=0\]

If $a=0$, it is a linear equation, so we may assume $a\not=0$.  Dividing by $a$
we get:

\[x^2+\frac{b}{a}x+\frac{c}{a}=0\]

We can make the substitution $x=y+k$, and we will choose k to eliminate the
linear term.

\[(y+k)^2+\frac{b}{a}(y+k)+\frac{c}{a}=0\]

\[y^2+2ky+k^2+\frac{b}{a}y+\frac{b}{a}k+\frac{c}{a}=0\]

\[y^2+(2k+\frac{b}{a})y+k^2+\frac{b}{a}k+\frac{c}{a}=0\]

From this, we see that to eliminate the linear term, we must choose
$k=\frac{-b}{2a}$.  Plugging in, we get:

\[y^2+\frac{b^2}{4a^2}-\frac{b^2}{2a^2}+\frac{c}{a}=0\]
\[y^2=\frac{b^2}{4a^2}-frac{c}{a}\]
\[y^2=\frac{b^2-4ac}{4a^2}\]
\[y=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\]
\[x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\]
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

There are a few observations here that will prove useful later.  The
first is the fact that there are, in general, two solutions, as long
as we include complex numbers in our solution set.  The fact that an
nth order polynomial equation has, in general, n solutions in the
complex numbers, is the {\em fundamental theorem of algebra}.

It is possible for $b^2-4ac$ to be $0$, in which case there is only one
solution.  We call this a {\em double root}.  The expression $b^2-4ac$
also determines whether the roots will be real or not.  This expression
is called the {\em discriminant} and has analogs in higher orders.

If we determine one root $\alpha_1$ of the $n$th order polynomial $f(x)$,
we can use polynomial division to factor $f$ into $(x-\alpha_1)f_1(x)$,
where $f_1$ is an $n-1$th order polynomial.  The fundamental theorem
of algebra says that we can continue to do so until we are left with a
linear equation.  This means that every $n$th order polynomial can be
written in {\em factored form} which looks like

\[a(x-\alpha_1)(x-\alpha_2)...(x-\alpha_n)\].

where $\alpha_i$ is a complex root.

There is also a certain symmetry between the two roots, in that if the
coefficients are real, if the roots are not real, they will come in
{\em conjugate pairs}, i.e. the real parts are the same and the imaginary
parts will be opposite in sign.  If the coefficients are rational, and
the roots are not, they will similarly come in pairs, so that they may both
be written as $a+b$ and $a-b$, where $a$ is rational and $b$ is the square
root of a non-square rational.

There are many generalizations of the last observation.  The first is that
the sum of the roots is always in the same field as the coefficients.  The
reason for this can be seen from the factored form of the equation,
multiplying out and only taking the $x^{n-1}$ term.  This is $-a\sum \alpha_i$.
So the sum of the roots is negative the coefficient of the $x^{n-1}$ term
divided by the coefficient of the $x^n$ term.

The second observation is that there is a ``conjugation'' map whenever
the roots are not in the field of coefficients.  For real coefficients and
non-real roots, this is the standard ``conjugation'' map in the complex
plane.  For rational coefficients and irrational roots, there is a similar
map that interchanges the roots.  A generalization of this idea is the
foundation of {\em Galois Theory}.

\section{The Cubic Equation}

To solve the cubic equation, we will begin by doing the same thing we
did in the quadratic equation, that is, first divide by the coefficient of
the cubic term to make it one, then translate to make the quadratic term
zero:

\[ax^3+bx^2+cx+d=0\]
\[x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a}=0\]
\[x=y+k\]

The quadratic term in y will then be:

\[3k+\frac{b}{a}\]

which we set to $0$ by $k=-\frac{b}{3a}$.

We now have the equation:

\[y^3+ey+f=0\]

The solution described by Cardano can be obtained by the surprising
substitution $y=A+B$, where $A$ and $B$ are yet to be determined.  We
then obtain:

\[A^3+3A^2B+3AB^2+B^3+e(A+B)+f=0\]
\[(A^3+B^3+f)+3AB(A+B)+e(A+B)=0\]
\[(A^3+B^3+f)+(3AB+e)(A+B)=0\]

We have the freedom to make both terms 0 by appropriate choices of $A$ and $B$,
so that:

\[A^3+B^3=-f\]
\[AB=-e/3\]

(We reject the solution $A+B=0$ in the second equation since then $A^3+B^3=0$,
and that is only true when $f=0$, in which case we can immediately factor
out one $y$, leaving a quadartic equation.)

\[B=-e/3A\]
\[A^3-\frac{e^3}{27A^3}=-f\]
\[(A^3)^2+fA^3-\frac{e^3}{27}=0\]

The last equation is a quadratic equation in $A^3$, which has, as roots:

\[A^3=\frac{-f\pm\sqrt{f^2-\frac{4e^3}{27}}}{2}\]
\[A^3=-\frac{f}{2}\pm\sqrt{(\frac{f}{2})^2-(\frac{e}{3})^3}\]
\[A=\sqrt[3]{-\frac{f}{2}\pm\sqrt{(\frac{f}{2})^2-(\frac{e}{3})^3}}\]

Since $A^3+B^3=-f$, if we choose one choice for the $\pm$ for $A$,
then $B$ will be the other root.  By symmetry between $A$ and $B$, we
do not get two different solutions by choosing $A$ to be one or the
other.  So we get one root as $A+B$.  The other roots can be determined
by factoring out $(y-A-B)$, and solving the quadratic, or by recognizing that
if the other two roots are $\alpha_2$ and $\alpha_3$, then by multiplying
out the factored form $(y-A-B)(y-\alpha_2)(y-\alpha_3)$, and comparing with
the original polynomial, we get:

\[\alpha_2+\alpha_3=-(A+B)\]
\[\alpha_2\alpha_3+(\alpha_2+\alpha_3)(A+B)=e\]
\[\alpha_2\alpha_3(A+B)=-f\]

Using the first equation in the second, we get:

\[\alpha_2\alpha_3=e+(A+B)^2=-f/(A+B)\]

so that $\alpha_2$ and $\alpha_3$ are roots of the equation:

\[z^2+(A+B)z+e+(A+B)^2=0\]

\[z=\frac{-(A+B)\pm\sqrt{(A+B)^2-4e-4(A+B)^2}}{2}\]

\[z=\frac{-(A+B)\pm\sqrt{-4e-3(A+B)^2}}{2}\]

Taking these three roots, we must then translate them by $\frac{b}{3a}$ to get
$x$.

\section{The Quartic Equation}

Since every non-real root to a polynomial with real coefficients is accompanied
by its complex conjugate, we can write every real quartic polynomial in terms
of two quadratic terms.  If a solution exists for this, the formula should
work for complex polynomials, since complex conjugation of non-real roots
is not used in the proof; only in the motivation.

Again, we can divide by the leading term and translate, completing the
quartic, to remove the cubic term, so we have:

\[x^4+ax^2+bx+c=0=(x^2+dx+e)(x^2+fx+g)\]

Multiplying out the right side, and determining the cubic term,
we find that $d=-f$, and that the following equations must be satisfied:

\begin{eqnarray}
e-d^2+g&=&a\\
d(g-e)&=&b\\
eg&=&c
\end{eqnarray}

Taking the first two equations, we have the sum and difference of $e$ and $g$,
which we can use to solve for $e$ and $g$.

\begin{eqnarray}
g+e&=&a+d^2\\
g-e&=&b/d
\end{eqnarray}

\begin{eqnarray}
g&=&(a+d^2+b/d)/2\\
e&=&(a+d^2-b/d)/2
\end{eqnarray}

To satisfy the third equation, we must have $d$ so that:

\[(a+d^2+b/d)(a+d^2-b/d)/4=c\]
\[(a+d^2)^2-(b/d)^2=4c\]
\[d^4+2ad^2+a^2-4c-b^2/d^2=0\]
\[d^6+2ad^4+(a^2-4c)d^2-b^2=0\]

which is a cubic equation in $d^2$.  If you find any root to this cubic
equation, and take either square root in $d$, you can use the above
equations to determine $g$ and $e$, and solve the quadratic equations
to get the roots to the original quartic.

\section{Approaches to the Quintic}

The Quintic cannot be written in terms of addition, subtraction,
multiplication, division, and radicals.  This was determined by Abel and
Galois and the generalization of the proof forms the branch of mathematics
called {\em Galois Theory}.  One can, however, solve this in terms of
the aforementioned functions and {\em hyperradicals}, which is the
real solution of the function $x^5+x-a=0$.  (This has exactly one root
since the derivative has no real roots.)

To do this, we divide by the coefficient of the quintic term, and apply
a certain transformation of the roots.  The kind of transformation we
will study is like a substitution but is only valid at the roots of the
polynomial.  The particular transformation is due to Tschirnhausen.

Let $x^5+ax^4+bx^3+cx^2+dx+e=0$.  We apply a transformation where
$y=\alpha x^4+\beta x^3+\gamma x^2+\delta x+\epsilon$ on the roots of the
original quintic.  There are infinitely many such transformations, even
when we know the coefficients, since we can, for instance, replace $x$ with
$-x^5-ax^4-bx^3-cx^2-e$ wherever it appears.  We will choose the
following:

\[\alpha x^4+\beta x^3+\gamma x^2+\delta x+(\epsilon-y)=0\]

Multiply by $x$ and replace $x^5$ with $-ax^4-bx^3-cx^2-dx-e$.
We now have a new quartic equation in $x$, which is linear in $y$, $\alpha$,
..., $\epsilon$.
We can do this four times, thereby obtaining five quartic equations in $x$
and linear in $y$, $\alpha$, ..., $\epsilon$.  We wish to find appropriate
$\alpha$,...,$\epsilon$ so that all these equations are satisfied, and this
is only the case if, when viewed as a linear transformation of
$(1,x,x^2,x^3,x^4)$, it is singular.  We can therefore write down the matrix
for this linear transformation, and require the determinant to be zero.

The determinant, when expanded, will be a homogeneous quintic equation
in the variables $y$, $\alpha$, ..., $\epsilon$.  The equation will thus
be $y^5+Q_1 y^4+Q_2 y^3 + Q_3 y^2+Q_4 y + Q_5=0$, where $Q_i$ is
a homogeneous $i$th order equation in $\alpha$, ..., $\epsilon$.

We can make the quartic, cubic, and quadratic terms zero by the following
method:

Solve $Q_1=0$ for $\epsilon$, we get $\epsilon$ as a linear function of
$\alpha$,...,$\delta$.  We can substitute this in $Q_2$ and $Q_3$,
so $Q_2$ is a homogeneous quadratic function of $\alpha$,...,$\delta$.
We now perform a linear transformation on these four coefficients so
that $Q_2$ is a sum of squares.  We can do this since if $\alpha^2$ is
non-zero, we can complete the square with the linear $\alpha$ terms,
and if it is zero, we can take a non-zero linear $\alpha$ term (say,
with $\beta$), and rotate 45 degrees in the $\alpha\beta$ plane,
generating an $\alpha^2$ term.  This can be continued with $\beta$ and
$gamma$ until $\delta^2$ is all that is left.

We can write $Q_2=u^2-v^2+s^2-t^2$, where $u,v,s,t$ are linear functions
of $\alpha, \beta, \gamma, \delta$.  Performing the inverse operation allows
us to write $Q_3$ in terms of these variables.  We can set $u=v$ and
$s=t$ to satisfy $Q_2$, and all we have left is a homogeneous cubic equation
in two variables (say, $s$ and $u$).

A homogeneous equation in two variables is equivalent to an inhomogeneous
equation of the same degree in one variable since solutions to the latter
can be determined by replacing $x^i$ terms with $x^iy^{n-i}$ terms, and
solving, then setting $y=1$; and solutions to the former can be determined
from the latter by setting $y=1$, solving the equation, thereby finding
all solutions of the form $(x,1)$, then multiplying the resulting $(x,1)$ by
a constant factor.

So instead of a homogeneous cubic equation in two variables, we have a
cubic equation in one variable, which we can solve.  Now, doing the
substitution, we have:

\[y^5+fy+g=0\]

We can now make a substitution $y=z\sqrt[4]{f}$, and we obtain:

\[z^5+z+g(f^{-\frac{5}{4}})=0\]

\section{Solution of the Cubic using Trigonometric Functions}

First, we observe that:

\[4\cos^3\theta-3\cos\theta-\cos(3\theta)=0\].

We transform the general cubic, again, to one of the form:

\[x^3+px+q=0\]

and substitute $x=m\cos\theta$.  We then have the equation:

\[m^3\cos^3\theta+pm\cos\theta+q=0\].

Then, comparing terms, we obtain:

\[\frac{m^3}{4}=\frac{pm}{-3}=\frac{q}{-\cos(3\theta)}\].

From this, we see that $m=\pm\sqrt{-\frac{4p}{3}}$ and
$\cos(3\theta)=\frac{3q}{pm}$.  The choice of sign in $m$ is not significant
since $-\cos 3\theta=\cos(3\pi-3\theta)=\cos(3(\pi-\theta))$, so that choosing
a different sign for $m$ only changes the sign of $\cos\theta$, and the
roots are of the form $m\cos\theta$.

The solutions, then, are: $m\cos(\theta)$, $m\cos(\theta+\pi/3)$, and
$m\cos(\theta+2\pi/3)$.

Note that if $p>0$, $m$ must be taken to be imaginary.  Now
$\cos(\frac{\pi}{2}+i\phi)=-i\sinh\phi$, so if $p>0$, 
$3\theta=\frac{\pi}{2}+i\phi$, and $\sinh\phi=\sqrt{\frac{4p}{3}}$, so
that $\theta=\frac{\pi}{6}+\frac{i}{3}\sinh^{-1}(\sqrt{\frac{4p}{3}})$.
The other roots can be obtained by adding $2\pi/3$ to $\theta$.

\end{document}
