\chapter{Polynomial in more than one variable}

\section{The Binonomial Theorem}

We know how to expand the binomial $(x+y)^2$.  Just play like $x$ and $y$ are
real numbers and multiply it out. 

\[  (x+y)^2 = (x+y)(x+y) = x(x+y) +y (x+y) = x x + xy + yx + yy = x^2 + 2xy +y^2 \]

In the same way, you can work out that the cube of the binomial $x+y$, $(x+y)^3$ can
be expanded to $x^3+3x^2y+3xy^2+y^3$.


 
\subsection{Pascal's Triangle}

\[
\begin{array}{ccc}
(x+y)^0  & = & 1\\
x+y & = & x+y\\
\left (x+y\right )^{2} & = & {x}^{2}+2\,xy+{y}^{2}\\
\left (x+y\right )^{3} & = & {x}^{3}+3\,{x}^{2}y+3\,x{y}^{2}+{y}^{3}\\
\left (x+y\right )^{4} & = & {x}^{4}+4\,{x}^{3}y+6\,{x}^{2}{y}^{2}+4\,x{y}^{
3}+{y}^{4}\\
\left (x+y\right )^{5} & = & {x}^{5}+5\,{x}^{4}y+10\,{x}^{3}{y}^{2}+10\,{x}^
{2}{y}^{3}+5\,x{y}^{4}+{y}^{5}\\
\left (x+y\right )^{6} & = & {x}^{6}+6\,{x}^{5}y+15\,{x}^{4}{y}^{2}+20\,{x}^
{3}{y}^{3}+15\,{x}^{2}{y}^{4}+6\,x{y}^{5}+{y}^{6} 
\end{array}
\]
  


\[
\begin{array}{ccc}
(x+y)^0  & = & 1\\
(x+y)^1 & = & x+y\\
\left (x+y\right )^{2} & = & {x}^{2}+2\,xy+{y}^{2}\\
\left (x+y\right )^{3} & = & {x}^{3}+3\,{x}^{2}y+3\,x{y}^{2}+{y}^{3}\\
\left (x+y\right )^{4} & = & {x}^{4}+4\,{x}^{3}y+6\,{x}^{2}{y}^{2}+4\,x{y}^{
3}+{y}^{4}\\
\cdots  & = & \cdots \\
\left (x+y\right )^{n} & = & {x}^{n}+n\,{x}^{n-1}y+\cdots + \binom{n}{i}\,{x}^
{n-i+1}{y}^{i-1}+\dots +6\,x{y}^{n-1}+{y}^{n} 
\end{array}
\]
  
\begin{theorem}[The Binomial Theorem]
Let $n$ and $i$ be integers with $n > 2$ and $1<i<n$. 
The coefficient of $x^iy^{n-i}$ in the expansion of $(x+y)^n$  is the sum of the
coefficients of $x^{i}y^{n-i-1}$ and $x^{i-1}y^{n-i}$ in the expansion of 
$(x+y)^{n-1}$.
\end{theorem}

{\bf Proof:}  We induct on $n$.
The statement is true if $n=3$ and $i=2$.  Suppose it is true for
$n-1$ and all $i$ between $1$ and $n-1$. 
We can write $(x+y)^{n } = (x+y)(x+y)^{n-1} = x(x+y)^{n-1}+y(x+y)^{n-1}$.
By induction, 

