Welcome to some respectable mathematics. Binomals look like this: $ (x + \text{something})(y + \text{something else} ) $. We are going to learn how to expand any variant of these, and then we will look at the special cases when $ x $ and $ y $ are the same and the \emph{something}'s are also the same; i.e. $ (x + a)(x + a)$. (There is a quick trick for solving these). Then we shall conclude the class with the second special case of the binomials - \emph{The Difference of Two Squares}. They come in the shape of $ (x + a)(x - a) $, and also can be easily expanded with a trick!

Before we get stuck in to the expansion tricks, let's make sure we understand what we are expanding.

What does the prefix \textit{\textbf{bi}} mean?
\mdots{2}

Examples of `\textbf{bi}' things include \dotfill\\\dotfill

Thus a \textbf{bi}nomial means \dotfill\\\dotfill

Now let us expand $ (a + 2)(b + 5) $. You just need to distribute each term in the first brackets with every term of the next set of brackets.

\vspace{0.5cm}
\begin{equation}
    (\source{a}+\source{2})(\target{b}+\target{5})= ab + 2b + 5a + 10\mbox{\drawarrows} 
\end{equation}
\vspace{0.5cm}

If at first you are struggling to remember the steps, just remember the acronym \textbf{FOIL}, \textbf{F}irst \textbf{O}utside \textbf{I}nside \textbf{L}ast.


Once again this has a geometric interpretation:

\begin{center}
\begin{tikzpicture}
    \draw (0,0) rectangle (6,6);
    \draw (0,0) rectangle (4,5);
    \draw (4,5) rectangle (6,6);

    \node at (2,6) [above] {$a$};
    \node at (0,5.5) [left] {$b$};
    \node at (0,2.5) [left] {$5$};
    \node at (5, 6) [above] {$2$};

    \node at (2,5.5) {$ab$};
    \node at (5,5.5) {$2b$};
    \node at (5,2.5) {$10$};
    \node at (2,2.5) {$5a$};

\end{tikzpicture}
\end{center}

And the area can now be computed by adding all the parts: \( ab + 2b + 5a + 10\), which is what our algebraic expansion told us too!

\newpage
\begin{examplebox}
    \subsection{Examples}
    \begin{enumerate}
        \item Expand the following:
            \begin{multicols}{2}
            \begin{enumerate}
                \item \((x+4)(x+5)=\)\mdots{2}
                \item \((x+3)(x-2)=\)\mdots{2}
                \item \((x-4)(x-3)=\)\mdots{2}
                \item \((2y+1)(3y-4)=\)\mdots{2}
            \end{enumerate}
            \end{multicols}
    \end{enumerate}
\end{examplebox}

\begin{exercisebox}
    \subsection{Exercises}
    \begin{multicols}{2}
    \begin{enumerate}[(a)]
        \item \((a+3)(a+9)=\)\mdots{2}
        \item \((a+8)(9+a)=\)\mdots{2}
        \item \((p-6)(p+4)=\)\mdots{2}
        \item \((x+3)(x-8)=\)\mdots{2}
        \item \((x+7)(x-4)=\)\mdots{2}
        \item \((5x+1)(x+2)=\)\mdots{2}
        \item \((4m+3)(2m-1)=\)\mdots{2}
        \item \((2x-7)(3x-1)=\)\mdots{2}
        \item \((2b+3)(4b-2)=\)\mdots{2}
        \item \((4c+d)(2c-3d)=\)\mdots{2}
        \item \((3x-y)(2x+5y)=\)\mdots{2}
        \item \((2p-5q)(3q-2p)=\)\mdots{2}
    \end{enumerate}
    \end{multicols}

\end{exercisebox}

