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