1. Problem: Factorize the polynomial $x^3 - x^2 - 17x - 15$ given it can be expressed as $(x + 3)(x^2 + bx + c)$. Find $b$ and $c$. Step 1: Expand the right-hand side: $$ (x + 3)(x^2 + bx + c) = x^3 + bx^2 + cx + 3x^2 + 3bx + 3c $$ Step 2: Combine like terms: $$ x^3 + (b + 3)x^2 + (c + 3b)x + 3c $$ Step 3: Equate to the left-hand side polynomial: Coefficients of $x^3$: $1 = 1$ (already matched) Coefficients of $x^2$: $b + 3 = -1$ Coefficients of $x$: $c + 3b = -17$ Constant term: $3c = -15$ Step 4: Solve the constant term equation: $$ 3c = -15 \Rightarrow c = -5 $$ Step 5: Use $c$ to find $b$ in the $x$ term equation: $$ -5 + 3b = -17 \Rightarrow 3b = -12 \Rightarrow b = -4 $$ Answer: $b = -4$, $c = -5$. --- 2. Problem: Fully factorize $x^3 - x^2 - 17x - 15$ using the found $b$ and $c$. Step 1: Substitute back the quadratic factor: $$ (x + 3)(x^2 - 4x - 5) $$ Step 2: Factorize the quadratic: $$ x^2 - 4x - 5 = (x - 5)(x + 1) $$ Final factorization: $$ (x + 3)(x - 5)(x + 1) $$ --- 3. Problem: Expand and simplify $(4 + \sqrt{3})(4 - \sqrt{3})$. Step 1: Use the difference of squares formula: $$ (4)^2 - (\sqrt{3})^2 = 16 - 3 = 13 $$ Answer: $13$. --- 4. Problem: Express $\frac{26}{4 + \sqrt{3}}$ in the form $a + b\sqrt{3}$, where $a$ and $b$ are integers. Step 1: Rationalize the denominator by multiplying numerator and denominator by $4 - \sqrt{3}$: $$ \frac{26}{4 + \sqrt{3}} \cdot \frac{4 - \sqrt{3}}{4 - \sqrt{3}} = \frac{26(4 - \sqrt{3})}{16 - 3} = \frac{26(4 - \sqrt{3})}{13} $$ Step 2: Simplify: $$ \frac{26}{13}(4 - \sqrt{3}) = 2(4 - \sqrt{3}) = 8 - 2\sqrt{3} $$ Answer: $8 - 2\sqrt{3}$. --- 5. Problem: Find $p$ such that $g(x) = x^2 + 3px + (14p - 3)$ has two equal roots. Step 1: For equal roots, discriminant $D = 0$. Step 2: Calculate discriminant: $$ D = (3p)^2 - 4(1)(14p - 3) = 9p^2 - 56p + 12 $$ Step 3: Set $D = 0$: $$ 9p^2 - 56p + 12 = 0 $$ Step 4: Use quadratic formula for $p$: $$ p = \frac{56 \pm \sqrt{(-56)^2 - 4 \cdot 9 \cdot 12}}{2 \cdot 9} = \frac{56 \pm \sqrt{3136 - 432}}{18} = \frac{56 \pm \sqrt{2704}}{18} = \frac{56 \pm 52}{18} $$ Step 5: Calculate two possibilities: $$ p = \frac{56 + 52}{18} = \frac{108}{18} = 6, \quad p = \frac{56 - 52}{18} = \frac{4}{18} = \frac{2}{9} $$ Only integer $p$ is $6$. Answer: $p = 6$. --- 6. Problem: For $p=6$, solve: $$ x^2 + 3 \times 6 x + (14 \times 6 - 3) = 0 $$ Step 1: Substitute: $$ x^2 + 18x + 81 = 0 $$ Step 2: Use quadratic formula: $$ x = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 81}}{2} = \frac{-18 \pm \sqrt{324 - 324}}{2} = \frac{-18}{2} = -9 $$ Since discriminant is zero, only one root: $x = -9$. Answer: $x = -9$. --- 7. Problem: Given height function of diver $h(t) = 5t - 10t^2 + 10$ for $t \geq 0$. (a) Find the height of springboard (height at $t=0$). $$ h(0) = 5(0) - 10(0)^2 + 10 = 10 $$ So, springboard is $10$ meters above water. (b) Find time diver hits water ($h(t)=0$): $$ 5t - 10t^2 + 10 = 0 \Rightarrow -10t^2 + 5t + 10 = 0 $$ Divide whole equation by $-5$: $$ 2t^2 - t - 2 = 0 $$ Use quadratic formula: $$ t = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 2 \times (-2)}}{2 \times 2} = \frac{1 \pm \sqrt{1 + 16}}{4} = \frac{1 \pm \sqrt{17}}{4} $$ Only positive root relevant: $$ t = \frac{1 + \sqrt{17}}{4} \approx 1.28 \text{ seconds} $$ (c) Express $h(t)$ in form $A - B(t - C)^2$: Start with: $$ h(t) = -10t^2 + 5t + 10 $$ Complete the square: $$ h(t) = -10\left(t^2 - \frac{1}{2}t \right) + 10 $$ Inside the bracket, complete the square: $$ t^2 - \frac{1}{2}t = \left(t - \frac{1}{4}\right)^2 - \frac{1}{16} $$ Thus, $$ h(t) = -10 \left[ \left(t - \frac{1}{4}\right)^2 - \frac{1}{16} \right] + 10 = -10 \left(t - \frac{1}{4}\right)^2 + \frac{10}{16} + 10 = -10 \left(t - \frac{1}{4}\right)^2 + 10.625 $$ Rewrite as: $$ h(t) = 10.625 - 10 \left(t - 0.25\right)^2 $$ Answer: $A=10.625$, $B=10$, $C=0.25$. (d) Maximum height is at vertex $t = C = 0.25$ seconds, height $A = 10.625$ meters. --- 8. Problem 5: Shade region satisfying inequalities: $$ y < x + 4, \quad y + 5x + 3 \geq 0, \quad y \geq -1, \quad x < 2 $$ These represent a polygon bounded by lines $y = x + 4$, $y = -5x - 3$, $y = -1$, and vertical line $x=2$. --- 9. Problem 6a: Solve inequality $$ 3(2x + 1) > 5 - 2x $$ Step 1: Expand and simplify: $$ 6x + 3 > 5 - 2x \Rightarrow 6x + 3 + 2x > 5 \Rightarrow 8x + 3 > 5 $$ $$ 8x > 2 \Rightarrow x > \frac{1}{4} $$ Answer: $x > \frac{1}{4}$. --- 10. Problem 6b: Solve inequality $$ 2x^2 - 7x + 3 > 0 $$ Step 1: Find roots of quadratic: $$ 2x^2 - 7x + 3 = 0 $$ Use quadratic formula: $$ x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm 5}{4} $$ Roots: $$ x=3 \text{ or } x=\frac{1}{2} $$ Step 2: Since $a=2 >0$, quadratic opens upwards; Inequality $>0$ true outside roots: $$ x < \frac{1}{2} \text{ or } x > 3 $$ Answer: $x < \frac{1}{2}$ or $x > 3$. --- 11. Problem 6c: Find $x$ satisfying both inequalities from 6a and 6b: $$ x > \frac{1}{4} \quad \text{and} \quad (x < \frac{1}{2} \text{ or } x > 3) $$ Step 1: Intersection: For $x > \frac{1}{4}$ and $x < \frac{1}{2}$, we get $\frac{1}{4} < x < \frac{1}{2}$. For $x > \frac{1}{4}$ and $x > 3$, we get $x > 3$. Answer: $$ \frac{1}{4} < x < \frac{1}{2} \text{ or } x > 3 $$ --- 12. Problem 7a: Find $x$ where $$ f(x) = \frac{2}{x}, \quad g(x) = 1, \quad f(x) < g(x), x \neq 0 $$ Inequality: $$ \frac{2}{x} < 1 $$ Step 1: Multiply both sides by $x$, considering $x>0$ and $x<0$ cases: - If $x > 0$: $$ 2 < x $$ - If $x < 0$: Multiply inequality reverses: $$ 2 > x $$ But since $x < 0$, $2 > x$ always true. Considering domain $x \neq 0$: Step 2: Combine: $$ x < 0 \quad \text{(since inequality holds)} \quad \text{or} \quad x > 2 $$ Answer: $$ x < 0 \text{ or } x > 2 $$ --- 13. Problem 7b: Find $x$ where $$ f(x) = \frac{2}{(x+1)^2}, \quad g(x) = 8 $$ Inequality: $$ \frac{2}{(x+1)^2} < 8, \quad x \neq -1 $$ Step 1: Multiply both sides by $(x+1)^2 > 0$ (square is always positive): $$ 2 < 8 (x+1)^2 $$ Step 2: Divide by 8: $$ \frac{2}{8} < (x+1)^2 \Rightarrow \frac{1}{4} < (x+1)^2 $$ Step 3: Take square root: $$ |x+1| > \frac{1}{2} $$ Step 4: Solve: $$ x + 1 > \frac{1}{2} \Rightarrow x > -\frac{1}{2} $$ $$ x + 1 < -\frac{1}{2} \Rightarrow x < -\frac{3}{2} = -1.5 $$ Answer: $$ x < -1.5 \text{ or } x > -0.5 $$ Since $x \neq -1$, these are valid. --- Summary answers: 1a) $b = -4$, $c = -5$ 1b) $(x+3)(x-5)(x+1)$ 2a) $13$ 2b) $8 - 2\sqrt{3}$ 3a) $p = 6$ 3b) $x = -9$ 4a) Height = $10$ 4b) Time = $1.28$ s 4c) $A=10.625$, $B=10$, $C=0.25$ 4d) Max height $= 10.625$ m at $0.25$s 5) Region bounded by $y < x+4$, $y \geq -5x -3$, $y \geq -1$, $x < 2$ 6a) $x > \frac{1}{4}$ 6b) $x < \frac{1}{2}$ or $x > 3$ 6c) $\frac{1}{4} < x < \frac{1}{2}$ or $x > 3$ 7a) $x < 0$ or $x > 2$ 7b) $x < -1.5$ or $x > -0.5$