We shall cover this one quickly so you have time to do a brick of exercises after :D. Difference of Two Square 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 ju 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}