\begin{questions} \Question[3] Evaluate \(2m(m-3n)\) when: \begin{multicols}{3} \begin{parts} \part \(m=3,n=5\) \begin{solutionordottedlines}[2cm] \(2 \cdot 3(3-3 \cdot 5) = 6(3-15) = 6(-12) = -72\) \end{solutionordottedlines} \part \(m=-3, n=-2\) \begin{solutionordottedlines}[2cm] \(2 \cdot (-3)(-3-3 \cdot (-2)) = -6(-3+6) = -6 \cdot 3 = -18\) \end{solutionordottedlines} \part \(m=\frac{1}{3}, n=\frac{1}{2}\) \begin{solutionordottedlines}[2cm] \(2 \cdot \frac{1}{3}\left(\frac{1}{3}-3 \cdot \frac{1}{2}\right) = \frac{2}{3}\left(\frac{1}{3}-\frac{3}{2}\right) = \frac{2}{3}\left(\frac{2-9}{6}\right) = \frac{2}{3} \cdot \frac{-7}{6} = -\frac{7}{9}\) \end{solutionordottedlines} \end{parts} \end{multicols} \Question[1] Evaluate \(\frac{p+2q}{3r}\) when \(p = 7, q = -2, r = 2\) \begin{solutionordottedlines}[2cm] \(\frac{7+2(-2)}{3 \cdot 2} = \frac{7-4}{6} = \frac{3}{6} = \frac{1}{2}\) \end{solutionordottedlines} \Question[1] Evaluate \(\frac{x+y}{3}\) when \(x = -6, y = -5\) \begin{solutionordottedlines}[2cm] \(\frac{-6-5}{3} = \frac{-11}{3}\) \end{solutionordottedlines} \Question[3] Fill in the missing term: \begin{multicols}{3} \begin{parts} \part \(2a+\dotfill=7a\) \begin{solutionordottedlines}[2cm] \(2a+5a=7a\) \end{solutionordottedlines} \part \(5m^2-\dotfill = -6m^2n\) \begin{solutionordottedlines}[2cm] \(5m^2-11m^2n = -6m^2n\) \end{solutionordottedlines} \part \(-6lm+\dotfill=lm\) \begin{solutionordottedlines}[2cm] \(-6lm+7lm=lm\) \end{solutionordottedlines} \end{parts} \end{multicols} \Question[6] Simplify by collecting like terms: \begin{multicols}{3} \begin{parts} \part \(9a^2+5a^2-12a^2\) \begin{solutionordottedlines}[2cm] \(9a^2+5a^2-12a^2 = 2a^2\) \end{solutionordottedlines} \part \(14a^2d-10a^2d-6a^2d\) \begin{solutionordottedlines}[2cm] \(14a^2d-10a^2d-6a^2d = -2a^2d\) \end{solutionordottedlines} \part \(17m^2-14m^2+8m^2\) \begin{solutionordottedlines}[2cm] \(17m^2-14m^2+8m^2 = 11m^2\) \end{solutionordottedlines} \part \(-4x^2+3x^2-3y-7y\) \begin{solutionordottedlines}[2cm] \(-4x^2+3x^2-3y-7y = -x^2-10y\) \end{solutionordottedlines} \part \(7x^3+6x^2-4y^3-x^2\) \begin{solutionordottedlines}[2cm] \(7x^3+6x^2-4y^3-x^2 = 7x^3+5x^2-4y^3\) \end{solutionordottedlines} \part \(-3ab^2+4a^2b-5ab^2+a^2b\) \begin{solutionordottedlines}[2cm] \(-3ab^2+4a^2b-5ab^2+a^2b = -8ab^2+5a^2b\) \end{solutionordottedlines} \end{parts} \end{multicols} \Question[12] Simplify \begin{multicols}{4} \begin{parts} \part \(4a\times 3b\) \begin{solutionordottedlines}[2cm] \(4a \times 3b = 12ab\) \end{solutionordottedlines} \part \(-2p \times(-3q)\) \begin{solutionordottedlines}[2cm] \(-2p \times(-3q) = 6pq\) \end{solutionordottedlines} \part \(27y \div 3\) \begin{solutionordottedlines}[2cm] \(27y \div 3 = 9y\) \end{solutionordottedlines} \part \(3 \times 12t \div 9\) \begin{solutionordottedlines}[2cm] \(3 \times 12t \div 9 = 4t\) \end{solutionordottedlines} \part \(24x\div 8 \times 3\) \begin{solutionordottedlines}[2cm] \(24x\div 8 \times 3 = 9x\) \end{solutionordottedlines} \part \(-\frac{12m}{18}\) \begin{solutionordottedlines}[2cm] \(-\frac{12m}{18} = -\frac{2}{3}m\) \end{solutionordottedlines} \part \(\frac{12ab}{4a}\) \begin{solutionordottedlines}[2cm] \(\frac{12ab}{4a} = 3b\) \end{solutionordottedlines} \part \(\frac{3x}{5}\times\frac{2}{3}\) \begin{solutionordottedlines}[2cm] \(\frac{3x}{5}\times\frac{2}{3} = \frac{2x}{5}\) \end{solutionordottedlines} \part \(\frac{2}{5a}\times\frac{1}{4a}\) \begin{solutionordottedlines}[2cm] \(\frac{2}{5a}\times\frac{1}{4a} = \frac{2}{20a^2} = \frac{1}{10a^2}\) \end{solutionordottedlines} \part \(\frac{3x}{5}\div\frac{3}{4}\) \begin{solutionordottedlines}[2cm] \(\frac{3x}{5}\div\frac{3}{4} = \frac{3x}{5} \times \frac{4}{3} = \frac{4x}{5}\) \end{solutionordottedlines} \part \(\frac{9y}{2}\div 18\) \begin{solutionordottedlines}[2cm] \(\frac{9y}{2}\div 18 = \frac{9y}{2} \times \frac{1}{18} = \frac{y}{4}\) \end{solutionordottedlines} \part \(\frac{5p}{6}\div(-\frac{10p}{3})\) \begin{solutionordottedlines}[2cm] \(\frac{5p}{6}\div(-\frac{10p}{3}) = \frac{5p}{6} \times \frac{-3}{10p} = -\frac{1}{4}\) \end{solutionordottedlines} \end{parts} \end{multicols} \Question[4] Fill in the missing boxes. Each box contains the product of the 2 boxes below it. \begin{multicols}{2} \begin{centering} \begin{parts} \part \begin{centering} \begin{tikzpicture} \draw (0,0) rectangle (1,1); \draw (1,0) rectangle (2,1); \draw (2,0) rectangle (3,1); \draw (0.5,1) rectangle (1.5,2); \draw (1.5,1) rectangle (2.5,2); \draw (1,2) rectangle (2,3); \node at (0.5,0.5) {$2a$}; \node at (1.5,0.5) {$4b$}; \node at (2.5,0.5) {$5a$}; \node at (1,1.5) {$8ab$}; \node at (2,1.5) {$20ab$}; \node at (1.5,2.5) {$160a^2b$}; \end{tikzpicture} \end{centering} \part \begin{centering} \begin{tikzpicture} \draw (0,0) rectangle (1,1); \draw (1,0) rectangle (2,1); \draw (2,0) rectangle (3,1); \draw (0.5,1) rectangle (1.5,2); \draw (1.5,1) rectangle (2.5,2); \draw (1,2) rectangle (2,3); \node at (1,1.5) {$24a$}; \node at (1.5,2.5) {$48a^2b$}; \node at (2.5,0.5) {$2b$}; \node at (0.5,0.5) {$12a$}; \node at (1.5,0.5) {$2a$}; \node at (2,1.5) {$4ab$}; \end{tikzpicture} \end{centering} \end{parts} \end{centering} \end{multicols} \Question[8] Expand: \begin{multicols}{4} \begin{parts} \part \(b(b+7)\) \begin{solutionordottedlines}[2cm] \(b(b+7) = b^2+7b\) \end{solutionordottedlines} \part \(4h(5h-7)\) \begin{solutionordottedlines}[2cm] \(4h(5h-7) = 20h^2-28h\) \end{solutionordottedlines} \part \(-k(5k-4)\) \begin{solutionordottedlines}[2cm] \(-k(5k-4) = -5k^2+4k\) \end{solutionordottedlines} \part \(-4x(3x-5)\) \begin{solutionordottedlines}[2cm] \(-4x(3x-5) = -12x^2+20x\) \end{solutionordottedlines} \part \(4c(2c-d)\) \begin{solutionordottedlines}[2cm] \(4c(2c-d) = 8c^2-4cd\) \end{solutionordottedlines} \part \(-3x(2x+5y)\) \begin{solutionordottedlines}[2cm] \(-3x(2x+5y) = -6x^2-15xy\) \end{solutionordottedlines} \part \(3p(2-5pq)\) \begin{solutionordottedlines}[2cm] \(3p(2-5pq) = 6p-15p^2q\) \end{solutionordottedlines} \part \(-10b(3a-7b)\) \begin{solutionordottedlines}[2cm] \(-10b(3a-7b) = -30ab+70b^2\) \end{solutionordottedlines} \end{parts} \end{multicols} \Question[6] Expand and collect like terms \begin{multicols}{3} \begin{parts} \part \(\frac{1}{4}(x+2)+\frac{x}{3}\) \begin{solutionordottedlines}[2cm] \(\frac{1}{4}(x+2)+\frac{x}{3} = \frac{1}{4}x+\frac{1}{2}+\frac{x}{3} = \frac{7}{12}x+\frac{1}{2}\) \end{solutionordottedlines} \part \(\frac{3}{7}(3x+5)+\frac{x}{3}\) \begin{solutionordottedlines}[2cm] \(\frac{3}{7}(3x+5)+\frac{x}{3} = \frac{9}{7}x+\frac{15}{7}+\frac{x}{3} = \frac{37}{21}x+\frac{15}{7}\) \end{solutionordottedlines} \part \(-\frac{1}{2}(3x+2)-\frac{2x}{5}\) \begin{solutionordottedlines}[2cm] \(-\frac{1}{2}(3x+2)-\frac{2x}{5} = -\frac{3}{2}x-1-\frac{2x}{5} = -\frac{19}{10}x-1\) \end{solutionordottedlines} \part \(2p(3p+1)-5(p+1)\) \begin{solutionordottedlines}[2cm] \(2p(3p+1)-5(p+1) = 6p^2+2p-5p-5 = 6p^2-3p-5\) \end{solutionordottedlines} \part \(2p(3p+1)-4(2p+1)\) \begin{solutionordottedlines}[2cm] \(2p(3p+1)-4(2p+1) = 6p^2+2p-8p-4 = 6p^2-6p-4\) \end{solutionordottedlines} \part \(4z(4z-2)-z(z+2)\) \begin{solutionordottedlines}[2cm] \(4z(4z-2)-z(z+2) = 16z^2-8z-z^2-2z = 15z^2-10z\) \end{solutionordottedlines} \end{parts} \end{multicols} \Question[8] Expand: \begin{multicols}{4} \begin{parts} \part \((x-6)(x-4)\) \begin{solutionordottedlines}[2cm] \(x^2 - 10x + 24\) \end{solutionordottedlines} \part \((4x+1)(3x-1)\) \begin{solutionordottedlines}[2cm] \(12x^2 - 1\) \end{solutionordottedlines} \part \((4x+3)(2x-1)\) \begin{solutionordottedlines}[2cm] \(8x^2 - 4x - 3\) \end{solutionordottedlines} \part \((x-4)(2x+5)\) \begin{solutionordottedlines}[2cm] \(2x^2 + x - 20\) \end{solutionordottedlines} \part \((x+3)(x+3)\) \begin{solutionordottedlines}[2cm] \(x^2 + 6x + 9\) \end{solutionordottedlines} \part \((2x-5)(x+3)\) \begin{solutionordottedlines}[2cm] \(2x^2 + x - 15\) \end{solutionordottedlines} \part \((2x+3)(2x+3)\) \begin{solutionordottedlines}[2cm] \(4x^2 + 12x + 9\) \end{solutionordottedlines} \part \((\frac{2b}{3}+2)(\frac{b}{5}-2)\) \begin{solutionordottedlines}[2cm] \(\frac{2b^2}{15} - \frac{4b}{3} - \frac{2b}{5} + 4\) \end{solutionordottedlines} \end{parts} \end{multicols} \Question[6] Fill in the blanks: \begin{multicols}{2} \begin{parts} \part \((x+5)(\fillin[x+3])=x^2+8x+15\) \part \((x+3)(\fillin[x-5])=x^2-2x-15\) \part \((3x+4)(\fillin[$\frac{1}{3}x-4$])=3x^2+x-4\) \part \((x+\fillin[3])(x+6)=x^2+9x+\fillin[18]\) \part \((2x+3)(\fillin[$\frac{1}{2}x+5$])=2x^2+7x+\fillin[15]\) \part \((\fillin[4] x - 3)(\fillin[-3] x 5 \fillin[-1]) = 12x^2-x-6\) \end{parts} \end{multicols} \Question[4] Expand \begin{multicols}{4} \begin{parts} \part \((x-7)^2\) \begin{solutionordottedlines}[2cm] \(x^2 - 14x + 49\) \end{solutionordottedlines} \part \((a+8)^8\) \begin{solutionordottedlines}[2cm] This is a binomial expansion that requires the binomial theorem to expand fully. \end{solutionordottedlines} \part \((9+x)^2\) \begin{solutionordottedlines}[2cm] \(x^2 + 18x + 81\) \end{solutionordottedlines} \part \(x-11)^2\) \begin{solutionordottedlines}[2cm] \(x^2 - 22x + 121\) \end{solutionordottedlines} \end{parts} \end{multicols} \Question[2] Expand \begin{multicols}{2} \begin{parts} \part \((\frac{2x}{5}-1)^2\) \begin{solutionordottedlines}[2cm] \(\frac{4x^2}{25} - \frac{4x}{5} + 1\) \end{solutionordottedlines} \part \((\frac{3x}{4}+\frac{2}{3})^2\) \begin{solutionordottedlines}[2cm] \(\frac{9x^2}{16} + 2x + \frac{4}{9}\) \end{solutionordottedlines} \end{parts} \end{multicols} \Question[3] Evaluate the following using $ (a+b)^2 = a^2 + 2ab + b^2 $ and $(a-b)^2 = a^2 - 2ab + b^2$. \begin{multicols}{3} \begin{parts} \part \((1.01)^2\) \begin{solutionordottedlines}[2cm] \(1.0201\) \end{solutionordottedlines} \part \((0.99)^2\) \begin{solutionordottedlines}[2cm] \(0.9801\) \end{solutionordottedlines} \part \((4.01)^2\) \begin{solutionordottedlines}[2cm] \(16.0801\) \end{solutionordottedlines} \end{parts} \end{multicols} \Question[4] Expand and collect like terms \begin{multicols}{2} \begin{parts} \part \((x-2)^2+(x-4)^2\) \begin{solutionordottedlines}[2cm] \(2x^2 - 12x + 20\) \end{solutionordottedlines} \part \((2x+5)^2+(2x-5)^2\) \begin{solutionordottedlines}[2cm] \(8x^2 + 50\) \end{solutionordottedlines} \part \(x^2+(x+1)^2+(x+2)^2+(x+3)^2\) \begin{solutionordottedlines}[2cm] \(4x^2 + 20x + 14\) \end{solutionordottedlines} \part \((\frac{x}{2}+1)^2 + (\frac{x}{2}-1)^2\) \begin{solutionordottedlines}[2cm] \(\frac{x^2}{2} + 2\) \end{solutionordottedlines} \end{parts} \end{multicols} \Question[6] Expand \begin{multicols}{2} \begin{parts} \part \((z-7)(z+7)\) \begin{solutionordottedlines}[2cm] \(z^2 - 49\) \end{solutionordottedlines} \part \((10-x)(10+x)\) \begin{solutionordottedlines}[2cm] \(100 - x^2\) \end{solutionordottedlines} \part \((3x-2)(3x+2)\) \begin{solutionordottedlines}[2cm] \(9x^2 - 4\) \end{solutionordottedlines} \part \((\frac{x}{2}+3)(\frac{x}{2}-3)\) \begin{solutionordottedlines}[2cm] \(\frac{x^2}{4} - 9\) \end{solutionordottedlines} \part \((\frac{x}{3}+\frac{1}{2})(\frac{x}{3}-\frac{1}{2})\) \begin{solutionordottedlines}[2cm] \(\frac{x^2}{9} - \frac{1}{4}\) \end{solutionordottedlines} \part Is $ a^2-2a+1 $ a perfect square expansion or a difference of 2 squares? \begin{solutionordottedlines}[2cm] Perfect square expansion of $(a-1)^2$. \end{solutionordottedlines} \end{parts} \end{multicols} \end{questions} \subsection{Challenge Problems} \begin{questions} \Question[5] \begin{parts} \part Show that the perimeter of the rectangle is \((4x+6)\)cm \begin{solutionordottedlines}[2cm] Perimeter = $2(AB + AD) = 2((x+3) + x) = 2(2x+3) = 4x+6$ cm. \end{solutionordottedlines} \part Find the perimeter if $AD = 2\text{cm}$ \begin{solutionordottedlines}[2cm] Perimeter = $2(AB + AD) = 2((2+3) + 2) = 2(5+2) = 14$ cm. \end{solutionordottedlines} \part Find $x$ if the perimeter = $36\text{cm}$ \begin{solutionordottedlines}[2cm] $4x+6 = 36 \Rightarrow 4x = 30 \Rightarrow x = 7.5$ cm. \end{solutionordottedlines} \part Find the area of $ABCD$ in terms of $x$ \begin{solutionordottedlines}[2cm] Area = $AB \times AD = (x+3) \times x = x^2 + 3x$ cm\(^2\). \end{solutionordottedlines} \part Find the area of the rectangle if $AB=6\text{cm}$ \begin{solutionordottedlines}[2cm] Area = $AB \times AD = 6 \times 2 = 12$ cm\(^2\). \end{solutionordottedlines} \end{parts} \begin{center} \begin{tikzpicture} \draw (0,0) rectangle (5,2); \draw (0,0) rectangle (0.2,0.2); \draw (4.8,0.2) rectangle (5,0); \draw (0,2) rectangle (0.2,1.8); \draw (5,2) rectangle (4.8,1.8); \node at (0,0) [below left] {$D$}; \node at (0,2) [above left] {$A$}; \node at (5,2) [above right] {$B$}; \node at (5,0) [below right] {$C$}; \node at (0,1) [left] {$x$}; \node at (2.5,2) [above] {$x+3$}; \end{tikzpicture} \end{center} \Question[6] Expand and collect: \begin{parts} \part \((x-1)(x^2+x+1)\) \begin{solutionordottedlines}[3cm] \(x^3 - 1\) \end{solutionordottedlines} \part \((x-1)(x^4+x^3+x^2+x+1)\) \begin{solutionordottedlines}[3cm] \(x^5 - 1\) \end{solutionordottedlines} \part What do you expect the result of expanding \((x-1)(x^9+x^8+\dots+1)\) will be? \begin{solutionordottedlines}[3cm] \(x^{10} - 1\) \end{solutionordottedlines} \end{parts} \end{questions}