1. Find the solution set of the following inequalities. (a) Solve $2x^2 + 5x + 2 > 0$. Step 1: Factor the quadratic expression. $$2x^2 + 5x + 2 = (2x + 1)(x + 2)$$ Step 2: Set each factor equal to zero to find critical points. $$2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$$ $$x + 2 = 0 \Rightarrow x = -2$$ Step 3: Determine intervals based on critical points: $(-\infty, -2)$, $(-2, -\frac{1}{2})$, and $(-\frac{1}{2}, \infty)$. Step 4: Test each interval in the inequality $(2x + 1)(x + 2) > 0$. - For $x < -2$, choose $x = -3$: $(2(-3)+1)(-3+2) = (-5)(-1) = 5 > 0$ (True) - For $-2 < x < -\frac{1}{2}$, choose $x = -1$: $(2(-1)+1)(-1+2) = (-1)(1) = -1 < 0$ (False) - For $x > -\frac{1}{2}$, choose $x = 0$: $(2(0)+1)(0+2) = (1)(2) = 2 > 0$ (True) Step 5: Solution set is $(-\infty, -2) \cup (-\frac{1}{2}, \infty)$. (b) Solve $(2 + x)(x - 5)(x + 1) > 0$. Step 1: Identify zeros: $x = -2, 5, -1$. Step 2: Order zeros: $-2 < -1 < 5$. Step 3: Test intervals: $(-\infty, -2)$, $(-2, -1)$, $(-1, 5)$, $(5, \infty)$. - $x = -3$: $(2-3)(-3-5)(-3+1) = (-1)(-8)(-2) = -16 < 0$ (False) - $x = -1.5$: $(2-1.5)(-1.5-5)(-1.5+1) = (0.5)(-6.5)(-0.5) = 1.625 > 0$ (True) - $x = 0$: $(2+0)(0-5)(0+1) = (2)(-5)(1) = -10 < 0$ (False) - $x = 6$: $(2+6)(6-5)(6+1) = (8)(1)(7) = 56 > 0$ (True) Step 4: Solution set is $(-2, -1) \cup (5, \infty)$. (c) Solve $\frac{2x - 4}{x + 3} > \frac{x + 2}{2x + 6}$. Step 1: Note $2x + 6 = 2(x + 3)$. Step 2: Rewrite inequality: $$\frac{2x - 4}{x + 3} > \frac{x + 2}{2(x + 3)}$$ Step 3: Multiply both sides by $2(x + 3)$, noting $x \neq -3$: $$2(x + 3) \times \frac{2x - 4}{x + 3} > 2(x + 3) \times \frac{x + 2}{2(x + 3)}$$ Simplify: $$2(2x - 4) > x + 2$$ Step 4: Expand and simplify: $$4x - 8 > x + 2$$ Step 5: Subtract $x$ from both sides: $$4x - 8 - x > 2$$ $$3x - 8 > 2$$ Step 6: Add 8 to both sides: $$3x > 10$$ Step 7: Divide both sides by 3: $$x > \frac{10}{3}$$ Step 8: Consider domain restriction $x \neq -3$. Step 9: Check sign of denominator to ensure inequality direction is preserved. Since $x > \frac{10}{3} > -3$, denominator $x + 3 > 0$, so inequality direction is correct. Step 10: Solution set is $\left(\frac{10}{3}, \infty\right)$. 2. For what values of $x$ is $\frac{2x - 4}{x + 3} < 0$? Step 1: Factor numerator: $$2x - 4 = 2(x - 2)$$ Step 2: The expression is negative when numerator and denominator have opposite signs. Step 3: Find zeros and undefined points: - Numerator zero at $x = 2$ - Denominator zero at $x = -3$ (excluded) Step 4: Test intervals: - $(-\infty, -3)$: choose $x = -4$ Numerator: $2(-4) - 4 = -12$ (negative) Denominator: $-4 + 3 = -1$ (negative) Fraction: negative/negative = positive (not < 0) - $(-3, 2)$: choose $x = 0$ Numerator: $2(0) - 4 = -4$ (negative) Denominator: $0 + 3 = 3$ (positive) Fraction: negative/positive = negative (true) - $(2, \infty)$: choose $x = 3$ Numerator: $2(3) - 4 = 2$ (positive) Denominator: $3 + 3 = 6$ (positive) Fraction: positive/positive = positive (not < 0) Step 5: Solution set is $(-3, 2)$. Final answers: (a) $(-\infty, -2) \cup (-\frac{1}{2}, \infty)$ (b) $(-2, -1) \cup (5, \infty)$ (c) $\left(\frac{10}{3}, \infty\right)$ 2. $(-3, 2)$