We shall cover this one quickly so you have time to do a brick of exercises after :D. Difference of Two Squares are the second special case of the \textbf{binomial} expansion, and come in the form $ (x + a)( x - a) $. Expanding this out with our usual \textbf{FOIL} method gives $ x^2 + \cancel{ax} - \cancel{ax} - a^2$ which just leaves $ x^2 - a^2$; how convenient!

This time I will leave the geometric intuition as an exercise, feel free to come to me before next class to explain your ideas!

 \begin{exercisebox}                                                                                            
     \subsection{Exercises}                                                                                     
     \begin{questions}                                                                                          
         \Question[3] Let's practise:                                                                           
             \begin{parts}                                                                                      
                 \part \((x-5)(x+5)=\)\fillin[$x^2 - 25$]{2}                                                      
                 \part \((3x-4)(3x+4)=\)\fillin[$9x^2 - 16$]{2}                                                   
                 \part \((a+b)(a-b)=\)\fillin[$a^2 - b^2$]{2}                                                     
             \end{parts}                                                                                        
         \Question[4] Now back to perfect squares:                                                              
             \begin{multicols}{2}                                                                               
             \begin{parts}                                                                                      
                 \part \((x+1)^2=\)\fillin[$x^2 + 2x + 1$]{2}                                                     
                 \part \((x+5)^2=\)\fillin[$x^2 + 10x + 25$]{2}                                                   
                 \part \((2+x)^2=\)\fillin[$x^2 + 4x + 4$]{2}                                                     
                 \part \((x+20)^2=\)\fillin[$x^2 + 40x + 400$]{2}                                                 
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
         \Question[6] Try a mix now:                                                                            
             \begin{multicols}{2}                                                                               
             \begin{parts}                                                                                      
                 \part \((3x-2)(3x+2)=\)\fillin[$9x^2 - 4$]{3}                                                    
                 \part \((3a-4b)^2=\)\fillin[$9a^2 - 24ab + 16b^2$]{2}                                            
                 \part \((2x+3y)^2=\)\fillin[$4x^2 + 12xy + 9y^2$]{2}                                             
                 \part \((5a+2b)(5a-2b)=\)\fillin[$25a^2 - 4b^2$]{2}                                              
                 \part \((\frac{x}{2}+3)^2=\)\fillin[$\frac{x^2}{4} + 3x + 9$]{2}                                 
                 \part \((3c-b)^2=\)\fillin[$9c^2 - 6bc + b^2$]{2}                                                
             \end{parts}                                                                                        
             \end{multicols}                                                                                    
     \end{questions}                                                                                            
 \end{exercisebox}                                                                                              
