1. Problem a: Solve the inequality $2x^2 + 5x + 2 > 0$. 2. Use the quadratic formula to find roots of $2x^2 + 5x + 2 = 0$: $$x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{-5 \pm \sqrt{25 - 16}}{4} = \frac{-5 \pm 3}{4}$$ 3. Roots are $x = \frac{-5 + 3}{4} = -\frac{1}{2}$ and $x = \frac{-5 - 3}{4} = -2$. 4. Since the leading coefficient $2 > 0$, the parabola opens upward, so $2x^2 + 5x + 2 > 0$ outside the roots. 5. Solution set: $x < -2$ or $x > -\frac{1}{2}$. 1. Problem b: Solve $(2 + x)(x - 5)(x + 1) > 0$. 2. Identify roots: $x = -2, 5, -1$. 3. Plot these on a number line and test intervals: - For $x < -2$, test $x = -3$: $(2 - 3)(-3 - 5)(-3 + 1) = (-1)(-8)(-2) = -16 < 0$. - For $-2 < x < -1$, test $x = -1.5$: $(2 - 1.5)(-1.5 - 5)(-1.5 + 1) = (0.5)(-6.5)(-0.5) = 1.625 > 0$. - For $-1 < x < 5$, test $x = 0$: $(2 + 0)(0 - 5)(0 + 1) = 2 \cdot (-5) \cdot 1 = -10 < 0$. - For $x > 5$, test $x = 6$: $(2 + 6)(6 - 5)(6 + 1) = 8 \cdot 1 \cdot 7 = 56 > 0$. 4. Solution set: $-2 < x < -1$ or $x > 5$. 1. Problem c: Solve $\frac{2x - 4}{x + 3} > \frac{x + 2}{2x + 6}$. 2. Note $2x + 6 = 2(x + 3)$, rewrite inequality: $$\frac{2x - 4}{x + 3} > \frac{x + 2}{2(x + 3)}$$ 3. Multiply both sides by $2(x + 3)$, noting $x \neq -3$: $$2(2x - 4) > x + 2$$ 4. Simplify: $$4x - 8 > x + 2$$ 5. Rearrange: $$4x - x > 2 + 8$$ $$3x > 10$$ $$x > \frac{10}{3}$$ 6. Check domain restrictions: $x \neq -3$ and denominator $2x + 6 \neq 0$. 7. Final solution: $x > \frac{10}{3}$.