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{enumerate}
        \item \((x-5)^2= \)\dotfill
        \item \((x+7)^2=\)\dotfill
        \item \((3x-1)^2=\)\dotfill
    \end{enumerate}
    \end{doublespace}
\end{examplebox}
