This is one of the special cases mentioned earlier. Our general binomial looks like $ (x + a)(y + b) $, but perfect squares are easier and look like $ (x + a)(x + a) $ which can then be simplified to be $ (x + a)^2 $.

Remember the trick mentioned earlier? This is it:
\begin{enumerate}
    \item Take the first term, square it
    \item Take the last term, square it
    \item Multiply all the terms with each other
\end{enumerate}

Thus we have $ (x + a)^2 = x^2 + 2ax + a^2 $. Simple as that.

Here is the geometric intuition:

\begin{center}
\begin{tikzpicture}
    \draw (0,0) rectangle (6,6);
    \draw (0,0) rectangle (4,2);
    \draw (4,2) rectangle (6,6);

    \node at (2,6) [above] {$x$};
    \node at (0,4) [left] {$x$};
    \node at (5,6) [above] {$a$};
    \node at (0,1) [left] {$a$};

    \node at (2,4) {$ x^2 $};
    \node at (5,4) {$ax$};
    \node at (5,1) {$a^2$};
    \node at (2,1) {$ax$};

\end{tikzpicture}
\end{center}

We take a square of $x$ units and extend it to a square of length $x+a$:

 \begin{examplebox}                                                                                             
     \subsection{Examples}                                                                                      
     Let's only do a few examples this time. We'll come back and do more practise after covering                
 \emph{differences of two squares.}                                                                             
     \begin{doublespace}                                                                                        
     \begin{questions}                                                                                          
         \question[]                                                                                            
         \begin{parts}                                                                                          
             \part \((x-5)^2= \)                                                                                
                 \begin{solutionordottedlines}[1cm]                                                             
                 $x^2 - 10x + 25$                                                                               
                 \end{solutionordottedlines}                                                                    
             \part \((x+7)^2=\)                                                                                 
                 \begin{solutionordottedlines}[1cm]                                                             
                 $x^2 + 14x + 49$                                                                               
                 \end{solutionordottedlines}                                                                    
             \part \((3x-1)^2=\)                                                                                
                 \begin{solutionordottedlines}[1cm]                                                             
                 $9x^2 - 6x + 1$                                                                                
                 \end{solutionordottedlines}                                                                    
         \end{parts}                                                                                            
     \end{questions}                                                                                            
     \end{doublespace}                                                                                          
 \end{examplebox}                                                                                               
