The expansion of a perfect square has a special form. For example:

\[
\begin{aligned}
(x+3)^{2} & =(x+3)(x+3) \\
& =x^{2}+6 x+9 \\
& =x^{2}+2 \times(3 x)+3^{2}
\end{aligned}
\]

\begin{theorembox}
    \subsection{Theorem}
    \begin{align}
        a^{2}+2 a b+b^{2}&=(a+b)^{2}\\
        a^{2}-2 a b+b^{2}&=(a-b)^{2}
    \end{align}
\end{theorembox}


\begin{examplebox}
    \subsection*{Examples:}
    \begin{questions}
        \question[3]
        \begin{parts}
            \part \(x^{2}+8 x+16\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+8 x+16=x^{2}+2 \times 4 x+4^{2}\)

                \[
                =(x+4)^{2}
                \]
            \end{solutionordottedlines}
            \part \(x^{2}-10 x+25\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-10 x+25=x^{2}-2 \times 5 x+5^{2}\)

                \[
                =(x-5)^{2}
                \]
            \end{solutionordottedlines}
            \part \(x^{2}+11 x+\frac{121}{4}\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+11 x+\frac{121}{4}=x^{2}+2 \times \frac{11}{2} x+\left(\frac{11}{2}\right)^{2}\)

                \[
                =\left(x+\frac{11}{2}\right)^{2}
                \]
            \end{solutionordottedlines}
        \end{parts}
        \question[2] Identify the simple quadratic expression that cannot be factorised as a perfect square.
        \begin{onehalfspacing}
        \begin{parts}\begin{multicols}{2}
            \part \(x^{2}+4 x+4\)
            \begin{solutionordottedlines}[0.5in]
                Can be factorised as \((x+2)^{2}\).
            \end{solutionordottedlines}
            \part \(x^{2}-6 x+12\)
            \begin{solutionordottedlines}[0.5in]
                Cannot be factorised as a perfect square because \(b^{2}-4ac = 36-48 < 0\).
            \end{solutionordottedlines}
            \part \(x^{2}-12 x+36\)
            \begin{solutionordottedlines}[0.5in]
                Can be factorised as \((x-6)^{2}\).
            \end{solutionordottedlines}
            \part \(x^{2}-10 x+25\)
            \begin{solutionordottedlines}[0.5in]
                Can be factorised as \((x-5)^{2}\).
            \end{solutionordottedlines}
            \part \(x^{2}+6 x+9\)
            \begin{solutionordottedlines}[0.5in]
                Can be factorised as \((x+3)^{2}\).
            \end{solutionordottedlines}
            \part \(x^{2}+5 x+\frac{25}{4}\)
            \begin{solutionordottedlines}[0.5in]
                Can be factorised as \(\left(x+\frac{5}{2}\right)^{2}\).
            \end{solutionordottedlines}
            \part \(x^{2}-8 x-16\)
            \begin{solutionordottedlines}[0.5in]
                Cannot be factorised as a perfect square because \(b^{2}-4ac = 64+64 > 0\) and is not a perfect square.
            \end{solutionordottedlines}
            \part \(x^{2}-14 x+49\)
            \begin{solutionordottedlines}[0.5in]
                Can be factorised as \((x-7)^{2}\).
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \end{onehalfspacing}
    \end{questions}
\end{examplebox}

\begin{exercisebox}
    \subsection*{Exercises:}
    \begin{questions}
        \question[8]
        \begin{parts}\begin{multicols}{2}
            \part \(x^{2}+12 x+36\)
            \begin{solutionordottedlines}[0.5in]
                \(x^{2}+12 x+36=(x+6)^{2}\)
            \end{solutionordottedlines}
            \part \(a^{2}-4 a+4\)
            \begin{solutionordottedlines}[0.5in]
                \(a^{2}-4 a+4=(a-2)^{2}\)
            \end{solutionordottedlines}
            \part \(x^{2}-9 x+\frac{81}{4}\)
            \begin{solutionordottedlines}[0.5in]
                \(x^{2}-9 x+\frac{81}{4}=\left(x-\frac{9}{2}\right)^{2}\)
            \end{solutionordottedlines}
            \part \(x^{2}+8 x+16=(x+\ldots)^{\ldots}\)
            \begin{solutionordottedlines}[0.5in]
                \(x^{2}+8 x+16=(x+4)^{2}\)
            \end{solutionordottedlines}
            \part \(x^{2}-\ldots+\ldots=(x-9)^{2}\)
            \begin{solutionordottedlines}[0.5in]
                \(x^{2}-18 x+81=(x-9)^{2}\)
            \end{solutionordottedlines}
            \part \(x^{2}-8 x+16\)
            \begin{solutionordottedlines}[0.5in]
                \(x^{2}-8 x+16=(x-4)^{2}\)
            \end{solutionordottedlines}
            \part \(m^{2}-26 m+169\)
            \begin{solutionordottedlines}[0.5in]
                \(m^{2}-26 m+169=(m-13)^{2}\)
            \end{solutionordottedlines}
            \part \(x^{2}+13 x+\frac{169}{4}\)
            \begin{solutionordottedlines}[0.5in]
                \(x^{2}+13 x+\frac{169}{4}=\left(x+\frac{13}{2}\right)^{2}\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \question[4] A brick company provides rectangular and square pavers.
        \begin{parts}
            \part{} Draw a diagram to show how two different square pavers of side lengths \(a\) and \(b\) respectively, and two identical rectangular pavers with dimensions \(a \times b\), can be arranged into a square.
            \begin{solutionordottedlines}[2in]
            \end{solutionordottedlines}
            \part{} How many of each type of paver enables you to pave a square area of side length \(a+3 b\)? Draw a diagram to illustrate how this can be done.
            \begin{solutionordottedlines}[2in]
                You would need 1 square paver of side length \(a\), 9 square pavers of side length \(b\), and 6 rectangular pavers of dimensions \(a \times b\).
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{exercisebox}