1. **Determine if** $f(x) = \left(\frac{3}{2}x^2 + \frac{1}{4}x - 5\right) + \left(3x^2 - \frac{5}{2}x + 6\right)$ **and** $g(x) = \left(\frac{5}{2}x^2 - \frac{3}{8}x - \frac{1}{4}\right) + \left(2x^2 - \frac{15}{8}x + \frac{5}{4}\right)$ **are equivalent.** 2. **Simplify each function by combining like terms:** $$f(x) = \left(\frac{3}{2}x^2 + \frac{1}{4}x - 5\right) + \left(3x^2 - \frac{5}{2}x + 6\right) = \left(\frac{3}{2}x^2 + 3x^2\right) + \left(\frac{1}{4}x - \frac{5}{2}x\right) + (-5 + 6)$$ $$= \frac{3}{2}x^2 + 3x^2 = \frac{3}{2}x^2 + \frac{6}{2}x^2 = \frac{9}{2}x^2$$ $$\frac{1}{4}x - \frac{5}{2}x = \frac{1}{4}x - \frac{10}{4}x = -\frac{9}{4}x$$ $$-5 + 6 = 1$$ So, $$f(x) = \frac{9}{2}x^2 - \frac{9}{4}x + 1$$ 3. **Simplify** $g(x)$: $$g(x) = \left(\frac{5}{2}x^2 - \frac{3}{8}x - \frac{1}{4}\right) + \left(2x^2 - \frac{15}{8}x + \frac{5}{4}\right) = \left(\frac{5}{2}x^2 + 2x^2\right) + \left(-\frac{3}{8}x - \frac{15}{8}x\right) + \left(-\frac{1}{4} + \frac{5}{4}\right)$$ $$\frac{5}{2}x^2 + 2x^2 = \frac{5}{2}x^2 + \frac{4}{2}x^2 = \frac{9}{2}x^2$$ $$-\frac{3}{8}x - \frac{15}{8}x = -\frac{18}{8}x = -\frac{9}{4}x$$ $$-\frac{1}{4} + \frac{5}{4} = \frac{4}{4} = 1$$ So, $$g(x) = \frac{9}{2}x^2 - \frac{9}{4}x + 1$$ 4. **Conclusion:** Since $f(x) = g(x)$, the functions are equivalent. --- 5. **Simplify and state restrictions:** a) Simplify $$\frac{12x^2 + 5x - 3}{16x^2 + 24x + 9}$$ Factor numerator: $$12x^2 + 5x - 3 = (3x - 1)(4x + 3)$$ Factor denominator: $$16x^2 + 24x + 9 = (4x + 3)^2$$ Simplify fraction: $$\frac{(3x - 1)(4x + 3)}{(4x + 3)(4x + 3)} = \frac{3x - 1}{4x + 3}$$ Restrictions: denominator $\neq 0 \Rightarrow 4x + 3 \neq 0 \Rightarrow x \neq -\frac{3}{4}$ b) Simplify $$\frac{x^4 + x^3 - 12x^2}{x^3 - 16x}$$ Factor numerator: $$x^2(x^2 + x - 12) = x^2(x + 4)(x - 3)$$ Factor denominator: $$x(x^2 - 16) = x(x - 4)(x + 4)$$ Simplify fraction: $$\frac{x^2(x + 4)(x - 3)}{x(x - 4)(x + 4)} = \frac{\cancel{x} x (x + 4)(x - 3)}{\cancel{x} (x - 4)(x + 4)} = \frac{x(x - 3)}{x - 4}$$ Restrictions: denominator $\neq 0 \Rightarrow x \neq 0, 4, -4$ --- 6. **Simplify and state restrictions:** Expression: $$\frac{4x^2 + 12x + 9}{x^2 - 25} \times \frac{x^2 - 4x - 5}{2x^2 - x - 6} + \frac{x^2 + 3x}{x - 2}$$ Factor each part: $$4x^2 + 12x + 9 = (2x + 3)^2$$ $$x^2 - 25 = (x - 5)(x + 5)$$ $$x^2 - 4x - 5 = (x - 5)(x + 1)$$ $$2x^2 - x - 6 = (2x + 3)(x - 2)$$ Rewrite expression: $$\frac{(2x + 3)^2}{(x - 5)(x + 5)} \times \frac{(x - 5)(x + 1)}{(2x + 3)(x - 2)} + \frac{x(x + 3)}{x - 2}$$ Simplify multiplication: $$\frac{(2x + 3)^2}{(x - 5)(x + 5)} \times \frac{(x - 5)(x + 1)}{(2x + 3)(x - 2)} = \frac{(2x + 3)^2 (x - 5)(x + 1)}{(x - 5)(x + 5)(2x + 3)(x - 2)}$$ Cancel common factors: $$= \frac{(2x + 3) \cancel{(2x + 3)} \cancel{(x - 5)} (x + 1)}{\cancel{(x - 5)} (x + 5) \cancel{(2x + 3)} (x - 2)} = \frac{(2x + 3)(x + 1)}{(x + 5)(x - 2)}$$ Add the remaining term: $$\frac{(2x + 3)(x + 1)}{(x + 5)(x - 2)} + \frac{x(x + 3)}{x - 2} = \frac{(2x + 3)(x + 1)}{(x + 5)(x - 2)} + \frac{x(x + 3)(x + 5)}{(x - 2)(x + 5)}$$ Common denominator $(x + 5)(x - 2)$: $$= \frac{(2x + 3)(x + 1) + x(x + 3)(x + 5)}{(x + 5)(x - 2)}$$ Expand numerator: $$(2x + 3)(x + 1) = 2x^2 + 2x + 3x + 3 = 2x^2 + 5x + 3$$ $$x(x + 3)(x + 5) = x(x^2 + 8x + 15) = x^3 + 8x^2 + 15x$$ Sum numerator: $$2x^2 + 5x + 3 + x^3 + 8x^2 + 15x = x^3 + (2x^2 + 8x^2) + (5x + 15x) + 3 = x^3 + 10x^2 + 20x + 3$$ Final simplified expression: $$\frac{x^3 + 10x^2 + 20x + 3}{(x + 5)(x - 2)}$$ Restrictions: denominators cannot be zero: $$x \neq 5, -5, 2$$ --- **Final answers:** - $f(x)$ and $g(x)$ are equivalent. - a) Simplified: $\frac{3x - 1}{4x + 3}$, $x \neq -\frac{3}{4}$ - b) Simplified: $\frac{x(x - 3)}{x - 4}$, $x \neq 0, 4, -4$ - c) Simplified: $\frac{x^3 + 10x^2 + 20x + 3}{(x + 5)(x - 2)}$, $x \neq 5, -5, 2$