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?
\begin{solutionordottedlines}[2cm]
The prefix \textit{\textbf{bi}} means two.                                                                     
\end{solutionordottedlines}

Examples of `\textbf{bi}' things include \fillin[binoculars (two eye pieces)] and \fillin[bicycle (two wheels)].

Thus a \textbf{bi}nomial means \fillin[a polynomial with 2 terms]

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{questions}                                                                                          
         \Question[4] Expand the following:                                                                     
             \begin{multicols}{2}                                                                               
             \begin{parts}                                                                                      
                 \part \((x+4)(x+5)=\)                                                                          
                 \begin{solutionordottedlines}[1cm]                                                             
                 $x^2 + 5x + 4x + 20 = x^2 + 9x + 20$                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \((x+3)(x-2)=\)                                                                          
                 \begin{solutionordottedlines}[1cm]                                                             
                 $x^2 - 2x + 3x - 6 = x^2 + x - 6$                                                              
                 \end{solutionordottedlines}                                                                    
                 \part \((x-4)(x-3)=\)                                                                          
                 \begin{solutionordottedlines}[1cm]                                                             
                 $x^2 - 3x - 4x + 12 = x^2 - 7x + 12$                                                           
                 \end{solutionordottedlines}                                                                    
                 \part \((2y+1)(3y-4)=\)                                                                        
                 \begin{solutionordottedlines}[1cm]                                                             
                 $6y^2 - 8y + 3y - 4 = 6y^2 - 5y - 4$                                                           
                 \end{solutionordottedlines}                                                                    
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
     \end{questions}                                                                                            
 \end{examplebox}                                                                                               

 \begin{exercisebox}                                                                                            
     \subsection{Exercises}                                                                                     
     \begin{questions}                                                                                          
         \question[]                                                                                            
         \begin{parts}\begin{multicols}{2}                                                                      
         \part \((a+3)(a+9)=\)                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                 
             $a^2 + 9a + 3a + 27 = a^2 + 12a + 27$                                                              
             \end{solutionordottedlines}                                                                        
         \part \((a+8)(9+a)=\)                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                 
             $a^2 + 9a + 8a + 72 = a^2 + 17a + 72$                                                              
             \end{solutionordottedlines}                                                                        
         \part \((p-6)(p+4)=\)                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                 
             $p^2 + 4p - 6p - 24 = p^2 - 2p - 24$                                                               
             \end{solutionordottedlines}                                                                        
         \part \((x+3)(x-8)=\)                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                 
             $x^2 - 8x + 3x - 24 = x^2 - 5x - 24$                                                               
             \end{solutionordottedlines}                                                                        
         \part \((x+7)(x-4)=\)                                                                                  
             \begin{solutionordottedlines}[2cm]                                                                 
             $x^2 - 4x + 7x - 28 = x^2 + 3x - 28$                                                               
             \end{solutionordottedlines}                                                                        
         \part \((5x+1)(x+2)=\)                                                                                 
             \begin{solutionordottedlines}[2cm]                                                                 
             $5x^2 + 10x + x + 2 = 5x^2 + 11x + 2$                                                              
             \end{solutionordottedlines}                                                                        
         \part \((4m+3)(2m-1)=\)                                                                                
             \begin{solutionordottedlines}[2cm]                                                                 
             $8m^2 - 4m + 6m - 3 = 8m^2 + 2m - 3$                                                               
             \end{solutionordottedlines}                                                                        
         \part \((2x-7)(3x-1)=\)                                                                                
             \begin{solutionordottedlines}[2cm]                                                                 
             $6x^2 - 2x - 21x + 7 = 6x^2 - 23x + 7$                                                             
             \end{solutionordottedlines}                                                                        
         \part \((2b+3)(4b-2)=\)                                                                                
             \begin{solutionordottedlines}[2cm]                                                                 
             $8b^2 - 4b + 12b - 6 = 8b^2 + 8b - 6$                                                              
             \end{solutionordottedlines}                                                                        
         \part \((4c+d)(2c-3d)=\)                                                                               
             \begin{solutionordottedlines}[2cm]                                                                 
             $8c^2 - 12cd + 2cd - 3d^2 = 8c^2 - 10cd - 3d^2$                                                    
             \end{solutionordottedlines}                                                                        
         \part \((3x-y)(2x+5y)=\)                                                                               
             \begin{solutionordottedlines}[2cm]                                                                 
             $6x^2 + 15xy - 2xy - 5y^2 = 6x^2 + 13xy - 5y^2$                                                    
             \end{solutionordottedlines}                                                                        
         \part \((2p-5q)(3q-2p)=\)                                                                              
             \begin{solutionordottedlines}[2cm]                                                                 
             $-4p^2 + 10pq + 6pq - 15q^2 = -4p^2 + 16pq - 15q^2$                                                
             \end{solutionordottedlines}                                                                        
         \end{multicols}\end{parts}                                                                             
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              

