We can factorise expressions entirely by realising that there is a some term which can be pulled out of all the terms. Consider \(4a + 12\). Here we have \fillin[3] common, and so this can be factorised as \fillin[$4(a+3)$]. Similarly, see for the following examples what term can be extracted from all the other ones.


\begin{examplebox}
    \subsection*{Examples:}
    \begin{questions}
        \question[8] Factorise: 
        \begin{parts}\begin{multicols}{2}
            \part \(12 x^{2}+3 x\)
            \begin{solutionordottedlines}[1in]
                \(12 x^{2}+3 x=3\left(4 x^{2}+x\right)\)
            \end{solutionordottedlines}
            \part \(36 a b-27 a\)
            \begin{solutionordottedlines}[1in]
                \(36 a b-27 a=9(4 a b-3 a)\)
                \(=3 x(4 x+1)\)
                \(=9 a(4 b-3)\)
            \end{solutionordottedlines}
            \part \(3 x+9\)
            \begin{solutionordottedlines}[1in]
                \(3 x+9=3(x+3)\)
            \end{solutionordottedlines}
            \part \(-7 a^{2}-49\)
            \begin{solutionordottedlines}[1in]
                \(-7 a^{2}-49=-7 \times a^{2}+(-7) \times 7\)
            \end{solutionordottedlines}
            \part \(7 a^{2}+63 a b\)
            \begin{solutionordottedlines}[1in]
                \(7 a^{2}+63 a b=7 a(a+9 b)\)
                \(=-7\left(a^{2}+7\right)\)
                \(7\left(-a^{2}-7\right)\) is also correct.
            \end{solutionordottedlines}
            \part \(5 p q^{2}+10 p^{2} q+25 p^{2} q^{2}\)
            \begin{solutionordottedlines}[1in]
                \(5 p q^{2}+10 p^{2} q+25 p^{2} q^{2}=5 p q(q+2 p+5 p q)\)
            \end{solutionordottedlines}
        \end{multicols}\end{parts}
    \end{questions}
\end{examplebox}

\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \begin{onehalfspacing}
        \question[8] Complete each factorisation.
        \begin{parts}\begin{multicols}{2}
            \part \(12 x=12 \times \fillin[x]\)
            \part \(24 a=12 \times \fillin[2a]\)
            \part \(15 a c=5 c \times \fillin[3a]\)
            \part \(y^{2}=y \times \fillin[y]\)
            \part \(24 a^{2}=6 a \times \fillin[4a]\)
            \part \(-6 b^{2}=2 b \times \fillin[-3b]\)
            \part \(4 x^{2} y=2 x \times \fillin[2xy]\)
            \part \(12 m^{2} n=3 m n \times \fillin[4m]\)
        \end{multicols}\end{parts}
        \question[4] Fill in the blanks by finding the missing factors.
            \begin{parts}\begin{multicols}{2}
            \part \(12 a+18=6 \times \fillin[(2a+3)]\)
            \part \(20 m n-15 n=5 n \times \fillin[(4m-3)]\)
            \part \(a^{2}+4 a=(a+4) \times \fillin[a]\)
            \part \(6 y z^{2}-18 y z=3 y z \times \fillin[(2z-6)]\)
            \end{multicols}\end{parts}
        \question[6] Factorise:
        \begin{parts}\begin{multicols}{2}
            \part \(6 x+24=\fillin[6(x+4)]\)
            \part \(5 a+15=\fillin[5(a+3)]\)
            \part \(y^{2}+x y=\fillin[y(y+x)]\)
            \part \(4 x+24=\fillin[4(x+6)]\)
            \part \(y^{2}-3 y=\fillin[y(y-3)]\)
            \part \(-14a - 21=\fillin[-7(2a+3)]\)
        \end{multicols}\end{parts}
        \question[8] Factorise:
        \begin{parts}\begin{multicols}{2}
            \part \(4 a b+16 a=\fillin[4a(b+4)]\)
            \part \(12 a^{2}+8 a=\fillin[4a(3a+2)]\)
            \part \(15 a^{2} b^{2}+10 a b^{2}=\fillin[$5ab^{2}(3a+2)$]\)
            \part \(4 a^{2}+6 a=\fillin[2a(2a+3)]\)
            \part \(3 b-6 b^{2}=\fillin[-3b(2b-1)]\)
            \part \(4 x^{2}-6 x y=\fillin[2x(2x-3y)]\)
            \part \(18 y-9 y^{2}=\fillin[9y(2-y)]\)
            \part \(4 a-6 a b^{2}=\fillin[$2a(2-3b^{2})$]\)
            \part \(14 m n^{2}-21 m^{2} n=\fillin[7mn(2n-3m)]\)
            \part \(6 p q^{2}-21 q p^{2}=\fillin[3pq(2q-7p)]\)
            \part \(-10 b^{2}+5 b=\fillin[5b(-2b+1)]\)
            \part \(-4 p q+16 p^{2}=\fillin[4p(-q+4p)]\)
            \part \(-16 a^{2} b-8 a b=\fillin[-8ab(2a+1)]\)
            \part \(-8 a^{2} b^{2}-2 a b=\fillin[-2ab(4ab+1)]\)
            \part \(-5 x^{2} y+30 x=\fillin[$5x(-x^{2}y+6)$]\)
            \part \(12 x y^{2}-3 x^{2} y=\fillin[3xy(4y-x)]\)
        \end{multicols}\end{parts}
        \question[3] In each part, an expression for the area of the rectangle has been given. Find an expression for the missing side length.
        \begin{parts}\begin{multicols}{3}
            \part
                \begin{tikzpicture}
                    \draw (0,0) -- (4,0) -- (4,2) -- (0,2) -- cycle;
                    \node at (2, 2) [above] {$2a + 3$};
                    \node at (2, 1) {Area = $8a + 12$};
                \end{tikzpicture}
                \begin{solutionordottedlines}[1in]
                    $4$
                \end{solutionordottedlines}
            \part
                \begin{tikzpicture}
                    \draw (0,0) -- (4,0) -- (4,2) -- (0,2) -- cycle;
                    \node at (2, 2) [above] {$5$};
                    \node at (2, 1) {Area = $10b + 15$};
                \end{tikzpicture}
                \begin{solutionordottedlines}[1in]
                    $2b+3$
                \end{solutionordottedlines}
            \part
                \begin{tikzpicture}
                    \draw (0,0) -- (4,0) -- (4,2) -- (0,2) -- cycle;
                    \node at (2, 2) [above] {$6a$};
                    \node at (2, 1) {Area = $12a^2 + 6ab$};
                \end{tikzpicture}
                \begin{solutionordottedlines}[1in]
                    $2a + b$
                \end{solutionordottedlines}
        \end{multicols}\end{parts}

        \question[4] Factorise:
        \begin{parts}\begin{multicols}{2}
            \part \(4 a^{2} b-2 a b+8 a b^{2}\)
                \begin{solutionordottedlines}[1in]
                    \(2ab(2a-1+4b)\)
                \end{solutionordottedlines}
            \part \(7 a b+14 a^{2}+21 b\)
                \begin{solutionordottedlines}[1in]
                    \(7(2a^{2}+ab+3b)\)
                \end{solutionordottedlines}
            \part \(5 a^{2} b+3 a b+4 a b^{2}\)
                \begin{solutionordottedlines}[1in]
                    \(ab(5a+3+4b)\)
                \end{solutionordottedlines}
            \part \(5 p^{2} q^{2}+10 p q^{2}+15 p^{2} q\)
                \begin{solutionordottedlines}[1in]
                    \(5pq(pq+2q+3p)\)
                \end{solutionordottedlines}
        \end{multicols}\end{parts}
      \end{onehalfspacing}
\end{questions}
\end{exercisebox}
