Chapter 6

1. $\begin{array}{rcl} 2 x {(2 x - 3)}^{2} & = & 2 x [{(2 x)}^{2} - 2 (2 x) (3) + 3^{2}] \\ = & 2 x (4 x^{2} - 12 x + 9) \\ = & 8 x^{3} - 24 x^{2} + 18 x \end{array}$ $\begin{array}{rcl} 2 x {(2 x - 3)}^{2} & = & 2 x [{(2 x)}^{2} - 2 (2 x) (3) + 3^{2}] \\ = & 2 x (4 x^{2} - 12 x + 9) \\ = & 8 x^{3} - 24 x^{2} + 18 x \end{array}$
2. $\begin{array}{l} \frac{1}{R} = \frac{1}{R_{1} + r} + \frac{1}{R_{2}} \\ \begin{array}{rcl} \frac{R R_{2} (R_{1} + r)}{R} & = & \frac{R R_{2} (R_{1} + r)}{R_{1} + r} + \frac{R R_{2} (R_{1} + r)}{R_{2}} \\ R_{2} (R_{1} + r) & = & R R_{2} + R (R_{1} + r) \\ R_{1} R_{2} + r R_{2} & = & R R_{2} + R R_{1} + r R \\ R_{1} R_{2} - R R_{1} & = & R R_{2} + r R - r R_{2} \\ R_{1} (R_{2} - R) & = & R R_{2} + r R - r R_{2} \\ R_{1} & = & \frac{R R_{2} + r R - r R_{2}}{R_{2} - R} \end{array} \end{array}$ $\begin{array}{l} \frac{1}{R} = \frac{1}{R_{1} + r} + \frac{1}{R_{2}} \\ \begin{array}{rcl} \frac{R R_{2} (R_{1} + r)}{R} & = & \frac{R R_{2} (R_{1} + r)}{R_{1} + r} + \frac{R R_{2} (R_{1} + r)}{R_{2}} \\ R_{2} (R_{1} + r) & = & R R_{2} + R (R_{1} + r) \\ R_{1} R_{2} + r R_{2} & = & R R_{2} + R R_{1} + r R \\ R_{1} R_{2} - R R_{1} & = & R R_{2} + r R - r R_{2} \\ R_{1} (R_{2} - R) & = & R R_{2} + r R - r R_{2} \\ R_{1} & = & \frac{R R_{2} + r R - r R_{2}}{R_{2} - R} \end{array} \end{array}$
3. $\frac{2 x^{2} + 5 x - 3}{2 x^{2} + 12 x + 18} = \frac{(2 x - 1) (x + 3)}{2 {(x + 3)}^{2}} = \frac{2 x - 1}{2 (x + 3)}$ $\frac{2 x^{2} + 5 x - 3}{2 x^{2} + 12 x + 18} = \frac{(2 x - 1) (x + 3)}{2 {(x + 3)}^{2}} = \frac{2 x - 1}{2 (x + 3)}$
4. $4 x^{2} - 16 y^{2} = 4 (x^{2} - 4 y^{2}) = 4 (x + 2 y) (x - 2 y)$ $4 x^{2} - 16 y^{2} = 4 (x^{2} - 4 y^{2}) = 4 (x + 2 y) (x - 2 y)$
5. $\begin{array}{rcl} p b^{3} + 8 a^{3} p & = & p (b^{3} + 8 a^{3}) \\ = & p (b + 2 a) (b^{2} - 2 a b + 4 a^{2}) \end{array}$ $\begin{array}{rcl} p b^{3} + 8 a^{3} p & = & p (b^{3} + 8 a^{3}) \\ = & p (b + 2 a) (b^{2} - 2 a b + 4 a^{2}) \end{array}$
6. $\begin{array}{rcl} 2 a - 4 T - b a + 2 b T & = & 2 (a - 2 T) - b (a - 2 T) \\ = & (2 - b) (a - 2 T) \end{array}$ $\begin{array}{rcl} 2 a - 4 T - b a + 2 b T & = & 2 (a - 2 T) - b (a - 2 T) \\ = & (2 - b) (a - 2 T) \end{array}$
7. $\begin{array}{rcl} 36 x^{2} + 14 x - 16 & = & 2 (18 x^{2} + 7 x - 8) \\ = & 2 (9 x + 8) (2 x - 1) \end{array}$ $\begin{array}{rcl} 36 x^{2} + 14 x - 16 & = & 2 (18 x^{2} + 7 x - 8) \\ = & 2 (9 x + 8) (2 x - 1) \end{array}$
8. $\begin{array}{rcl} \frac{3}{4 x^{2}} - \frac{2}{x^{2} - x} - \frac{x}{2 x - 2} & = & \frac{3}{4 x^{2}} - \frac{2}{x (x - 1)} - \frac{x}{2 (x - 1)} \\ = & \frac{3 (x - 1) - 2 (4 x) - x (2 x^{2})}{4 x^{2} (x - 1)} \\ = & \frac{3 x - 3 - 8 x - 2 x^{3}}{4 x^{2} (x - 1)} \\ = & \frac{- 2 x^{3} - 5 x - 3}{4 x^{2} (x - 1)} \end{array}$ $\begin{array}{rcl} \frac{3}{4 x^{2}} - \frac{2}{x^{2} - x} - \frac{x}{2 x - 2} & = & \frac{3}{4 x^{2}} - \frac{2}{x (x - 1)} - \frac{x}{2 (x - 1)} \\ = & \frac{3 (x - 1) - 2 (4 x) - x (2 x^{2})}{4 x^{2} (x - 1)} \\ = & \frac{3 x - 3 - 8 x - 2 x^{3}}{4 x^{2} (x - 1)} \\ = & \frac{- 2 x^{3} - 5 x - 3}{4 x^{2} (x - 1)} \end{array}$
9. $\begin{array}{rcl} \frac{x^{2} + x}{2 - x} \div \frac{x^{2}}{x^{2} - 4 x + 4} & = & \frac{x^{2} + x}{2 - x} \times \frac{x^{2} - 4 x + 4}{x^{2}} \\ = & \frac{x (x + 1)}{2 - x} \times \frac{{(x - 2)}^{2}}{x^{2}} \\ = & - \frac{x (x + 1) {(x - 2)}^{2}}{(x - 2) (x^{2})} \\ = & - \frac{(x + 1) (x - 2)}{x} \end{array}$ $\begin{array}{rcl} \frac{x^{2} + x}{2 - x} \div \frac{x^{2}}{x^{2} - 4 x + 4} & = & \frac{x^{2} + x}{2 - x} \times \frac{x^{2} - 4 x + 4}{x^{2}} \\ = & \frac{x (x + 1)}{2 - x} \times \frac{{(x - 2)}^{2}}{x^{2}} \\ = & - \frac{x (x + 1) {(x - 2)}^{2}}{(x - 2) (x^{2})} \\ = & - \frac{(x + 1) (x - 2)}{x} \end{array}$
10. $\begin{array}{rcl} \frac{1 - \frac{3}{2 x + 2}}{\frac{x}{5} - \frac{1}{2}} & = & \frac{\frac{2 (x + 1) - 3}{2 (x + 1)}}{\frac{2 x - 5}{10}} \\ = & \frac{2 x + 2 - 3}{2 (x + 1)} \times \frac{10}{2 x - 5} \\ = & \frac{5 (2 x - 1)}{(x + 1) (2 x - 5)} \end{array}$ $\begin{array}{rcl} \frac{1 - \frac{3}{2 x + 2}}{\frac{x}{5} - \frac{1}{2}} & = & \frac{\frac{2 (x + 1) - 3}{2 (x + 1)}}{\frac{2 x - 5}{10}} \\ = & \frac{2 x + 2 - 3}{2 (x + 1)} \times \frac{10}{2 x - 5} \\ = & \frac{5 (2 x - 1)}{(x + 1) (2 x - 5)} \end{array}$
11. $\begin{array}{l} Let t = time working together & \frac{48 t}{12} + \frac{48 t}{16} = 48 \\ \frac{t}{12} + \frac{t}{16} = 1 & 4 t + 3 t = 48 \\ LCD of 12 and 16 is 48. & t = \frac{48}{7} = 6.9 days \end{array}$ $\begin{array}{l} Let t = time working together & \frac{48 t}{12} + \frac{48 t}{16} = 48 \\ \frac{t}{12} + \frac{t}{16} = 1 & 4 t + 3 t = 48 \\ LCD of 12 and 16 is 48. & t = \frac{48}{7} = 6.9 days \end{array}$
12. $\begin{array}{l} \frac{3}{2 x^{2} - 3 x} + \frac{1}{x} = \frac{3}{2 x - 3} LCD = x (2 x - 3) \\ \frac{3 x (2 x - 3)}{x (2 x - 3)} + \frac{x (2 x - 3)}{x} = \frac{3 x (2 x - 3)}{2 x - 3} \\ \begin{array}{rcl} 3 + (2 x - 3) & = & 3 x \\ x & = & 0 \end{array} \end{array}$ $\begin{array}{l} \frac{3}{2 x^{2} - 3 x} + \frac{1}{x} = \frac{3}{2 x - 3} LCD = x (2 x - 3) \\ \frac{3 x (2 x - 3)}{x (2 x - 3)} + \frac{x (2 x - 3)}{x} = \frac{3 x (2 x - 3)}{2 x - 3} \\ \begin{array}{rcl} 3 + (2 x - 3) & = & 3 x \\ x & = & 0 \end{array} \end{array}$

No solution due to division by zero in first two terms

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Table of Contents for Chapter 6

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Table of Contents for
Chapter 6