The nomenclature ``simple'' quadratic means that the coefficient of $x^2$ is just 1. Let us recall some definitions:

\begin{questions}
    \question[1]
    \begin{boxdef}
        Nomenclature: \begin{solutionordottedlines}[1cm]the terminology for a specific topic\end{solutionordottedlines}
    \end{boxdef}
    \question[1]
    \begin{boxdef}
        Quadratic: \begin{solutionordottedlines}[1cm]a polynomial whose highest power is 2\end{solutionordottedlines}
    \end{boxdef}
    \question[1]
    \begin{boxdef}
        Coefficient: \begin{solutionordottedlines}[1cm]the number before the variable\end{solutionordottedlines}
    \end{boxdef}
\end{questions}

\fbox{\parbox{\textwidth}{\textbf{The Technique:}\\Let us take a concrete example: $x^2-3x-18$. To factorise this monic quadratic we need to find two numbers that multiply to give -18 and add to give -3. Such numbers will be -6 and 3. Thus we can write \(x^{2}-3 x-18=(x+3)(x-6)\)}}

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

                \[
                =(x+4)^{2}
                \]
            \end{solutionordottedlines}
            \part\(x^{2}+5 x+6\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+5 x+6=(x+2)(x+3)\)
            \end{solutionordottedlines}
            \part\(x^{2}+7 x+10\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+7 x+10=(x+2)(x+5)\)
            \end{solutionordottedlines}
            \part\(x^{2}+9 x+14\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+9 x+14=(x+2)(x+7)\)
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}
\end{examplebox}

\begin{exercisebox}
    \subsection*{Exercises:}
    \begin{questions}
        \question[4] Factorise:
        \begin{parts}\begin{multicols}{2}
            \part\(x^{2}+9 x+20\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+9 x+20=(x+4)(x+5)\)
            \end{solutionordottedlines}
            \part\(x^{2}+12 x+32\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+12 x+32=(x+4)(x+8)\)
            \end{solutionordottedlines}
            \part\(x^{2}+20 x+75\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+20 x+75=(x+5)(x+15)\)
            \end{solutionordottedlines}
            \part\( x^{2}+15 x+56\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+15 x+56=(x+7)(x+8)\)
            \end{solutionordottedlines}
            \part\(x^{2}-5 x+6\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-5 x+6=(x-2)(x-3)\)
            \end{solutionordottedlines}
            \part\(x^{2}-17 x+30\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-17 x+30=(x-2)(x-15)\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \begin{parts}\begin{multicols}{2}
            \setcounter{partno}{6}
            \part\(x^{2}-9 x+14\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-9 x+14=(x-2)(x-7)\)
            \end{solutionordottedlines}
            \part\(x^{2}-15 x+44\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-15 x+44=(x-4)(x-11)\)
            \end{solutionordottedlines}
            \part\(x^{2}-18 x+80\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-18 x+80=(x-10)(x-8)\)
            \end{solutionordottedlines}
            \part\(x^{2}-14 x+40\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-14 x+40=(x-4)(x-10)\)
            \end{solutionordottedlines}
            \part\(x^{2}-30 x+56\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-30 x+56=(x-8)(x-7)\)
            \end{solutionordottedlines}
            \part\(x^{2}+x-6\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+x-6=(x-2)(x+3)\)
            \end{solutionordottedlines}
            \part\(x^{2}+x-30\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+x-30=(x-5)(x+6)\)
            \end{solutionordottedlines}
            \part\(x^{2}-5 x-14\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-5 x-14=(x+2)(x-7)\)
            \end{solutionordottedlines}
            \part\(x^{2}-7 x-44\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-7 x-44=(x+4)(x-11)\)
            \end{solutionordottedlines}
            \part\(x^{2}+2 x-80\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+2 x-80=(x-8)(x+10)\)
            \end{solutionordottedlines}
            \part\(x^{2}+3 x-40\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+3 x-40=(x-5)(x+8)\)
            \end{solutionordottedlines}
            \part\(x^{2}+4 x-21\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+4 x-21=(x-3)(x+7)\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
        \begin{parts}\begin{multicols}{2}
            \setcounter{partno}{12}
            \part\(x^{2}-x-56\)
            \begin{solutionordottedlines}[1in]
                \((x-8)(x+7)\)
            \end{solutionordottedlines}
            \part\(x^{2}-3 x+2\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-3 x+2=(x-1)(x-2)\)
            \end{solutionordottedlines}
            \part\(x^{2}-3 x-10\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-3 x-10=(x+2)(x-5)\)
            \end{solutionordottedlines}
            \part\(x^{2}-5 x-14\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-5 x-14=(x+2)(x-7)\)
            \end{solutionordottedlines}
            \part\(x^{2}-5 x+4\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-5 x+4=(x-1)(x-4)\)
            \end{solutionordottedlines}
            \part\(x^{2}-x-12\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}-x-12=(x+3)(x-4)\)
            \end{solutionordottedlines}
            \part\(x^{2}+3 x-10\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+3 x-10=(x-2)(x+5)\)
            \end{solutionordottedlines}
            \part\(x^{2}+6 x+9\)
            \begin{solutionordottedlines}[1in]
                \(x^{2}+6 x+9=(x+3)(x+3)\)

                \[
                =(x+3)^{2}
                \]
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \end{questions}
\end{exercisebox}
