Welcome to the last section for today: Multiplication and Division of Surds. It is exactly what you would expect:

\begin{equation}
    \source{4}\source{\sqrt{7}}\times\target{2}\target{\sqrt{2}}= 8\sqrt{14}\mbox{\drawarrowsodd} 
\end{equation}

Realise that you do not need the terms to be \emph{alike} to multiply and divide them.

 \begin{examplebox}                                                                                                                          
     \section{Examples}                                                                                                                     
     \begin{multicols}{2}                                                                                                                    
         \begin{questions}                                                                                                                   
             \Question[2] $\sqrt{3}\times \sqrt{11}=$                                                                                        
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $\sqrt{33}$                                                                                                                 
                 \end{solutionordottedlines}                                                                                                 
             \question $\sqrt{15}\div\sqrt{3}=$                                                                                              
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $\sqrt{5}$                                                                                                                  
                 \end{solutionordottedlines}                                                                                                 
             \Question[2] $3\times 7\sqrt{5}=$                                                                                               
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $21\sqrt{5}$                                                                                                                
                 \end{solutionordottedlines}                                                                                                 
             \question $\sqrt{2}\times 4=$                                                                                                   
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $4\sqrt{2}$                                                                                                                 
                 \end{solutionordottedlines}                                                                                                 
             \Question[2] $\sqrt{6}\times\sqrt{6}=$                                                                                          
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $6$                                                                                                                         
                 \end{solutionordottedlines}                                                                                                 
             \question $(2\sqrt{6})^2=$                                                                                                      
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $24$                                                                                                                        
                 \end{solutionordottedlines}                                                                                                 
             \Question[2] $\sqrt{42}\div\sqrt{6}=$                                                                                           
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $\sqrt{7}$                                                                                                                  
                 \end{solutionordottedlines}                                                                                                 
             \question $\sqrt{35}\div\sqrt{10}=$                                                                                             
                 \begin{solutionordottedlines}[1cm]                                                                                          
                 $\sqrt{3.5}$ or $\sqrt{\frac{7}{2}}$                                                                                        
                 \end{solutionordottedlines}                                                                                                 
         \end{questions}                                                                                                                     
     \end{multicols}                                                                                                                         
 \end{examplebox}                                                                                                                            

 \begin{exercisebox}                                                                                                                         
     \section{Exercises}                                                                                                                    
     The distributive law is exactly the same as last week:                                                                                  
     \begin{equation}                                                                                                                        
         (\source{a}+\source{b})(\target{c}+\target{d})= ac + ad + bc + bd\mbox{\drawarrows}                                                 
     \end{equation}                                                                                                                          
                                                                                                                                             
     Expand and simplify the following:                                                                                                      
     \begin{questions}                                                                                                                       
         \Question[1] $2\sqrt{3}(4+3\sqrt3)=$                                                                                                
             \begin{solutionordottedlines}[2cm]                                                                                              
             $8\sqrt{3}+18$                                                                                                                  
             \end{solutionordottedlines}                                                                                                     
         \Question[1] $(3\sqrt{7}+1)(5\sqrt{7}-4)=$                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                              
             $15\sqrt{49} - 12\sqrt{7} + 5\sqrt{7} - 4 = 105 + 5\sqrt{7} - 12\sqrt{7} - 4 = 101 - 7\sqrt{7}$                                 
             \end{solutionordottedlines}                                                                                                     
         \Question[1] $(5\sqrt{2}-3)(2\sqrt{2}-4)=$                                                                                          
             \begin{solutionordottedlines}[2cm]                                                                                              
             $10\sqrt{4} - 20\sqrt{2} - 6\sqrt{2} + 12 = 20 - 26\sqrt{2} + 12 = 32 - 26\sqrt{2}$                                             
             \end{solutionordottedlines}                                                                                                     
         \Question[1] $(3\sqrt{2}-4\sqrt{3})(5\sqrt{3}-\sqrt{2})=$                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                              
             $15\sqrt{6} - 3\sqrt{4} - 20\sqrt{9} + 4\sqrt{6} = 19\sqrt{6} - 6 - 60 = -66 + 19\sqrt{6}$                                      
             \end{solutionordottedlines}                                                                                                     
         \Question[1] $(1-\sqrt{2})(3+2\sqrt{2})=$                                                                                           
             \begin{solutionordottedlines}[2cm]                                                                                              
             $3 + 2\sqrt{4} - 3\sqrt{2} - 2\sqrt{4} = 3 + 4 - 3\sqrt{2} - 4 = -3\sqrt{2} + 3$                                                
             \end{solutionordottedlines}                                                                                                     
     \end{questions}                                                                                                                         
                                                                                                                                             
 \end{exercisebox}                                                                                                                           
