1. Stating the problem: Simplify the expression (a) $$\left(\frac{3}{2}x^2 - \frac{1}{3}x + 2\right) + \frac{2}{3}x\left(-\frac{3}{2}x + 4 - \frac{1}{2}x^2\right)$$ 2. Use distributive property to expand the second term: $$\frac{2}{3}x \times -\frac{3}{2}x = \frac{2}{3} \times -\frac{3}{2} x \times x = \cancel{\frac{2}{3}} \times \cancel{-\frac{3}{2}} x^2 = -1 x^2 = -x^2$$ $$\frac{2}{3}x \times 4 = \frac{8}{3} x$$ $$\frac{2}{3}x \times -\frac{1}{2}x^2 = -\frac{2}{3} \times \frac{1}{2} x^3 = -\frac{1}{3} x^3$$ 3. Rewrite the expression: $$\frac{3}{2}x^2 - \frac{1}{3}x + 2 - x^2 + \frac{8}{3}x - \frac{1}{3}x^3$$ 4. Combine like terms: For $x^3$: only $-\frac{1}{3}x^3$ For $x^2$: $\frac{3}{2}x^2 - x^2 = \frac{3}{2}x^2 - \frac{2}{2}x^2 = \frac{1}{2}x^2$ For $x$: $-\frac{1}{3}x + \frac{8}{3}x = \frac{7}{3}x$ Constant: $2$ 5. Final simplified expression for (a): $$-\frac{1}{3}x^3 + \frac{1}{2}x^2 + \frac{7}{3}x + 2$$ --- 1. Stating the problem: Simplify the expression (b) $$\left(\frac{1}{2} x^{2} - \frac{3}{4} x + 1\right) \left(-\frac{2}{3} x + \frac{1}{2}\right) - \left(\frac{1}{3} x^{2} - x\right)$$ 2. Expand the product using distributive property: $$\left(\frac{1}{2} x^{2}\right) \left(-\frac{2}{3} x\right) = -\frac{1}{3} x^{3}$$ $$\left(\frac{1}{2} x^{2}\right) \left(\frac{1}{2}\right) = \frac{1}{4} x^{2}$$ $$\left(-\frac{3}{4} x\right) \left(-\frac{2}{3} x\right) = \frac{1}{2} x^{2}$$ $$\left(-\frac{3}{4} x\right) \left(\frac{1}{2}\right) = -\frac{3}{8} x$$ $$1 \times -\frac{2}{3} x = -\frac{2}{3} x$$ $$1 \times \frac{1}{2} = \frac{1}{2}$$ 3. Sum all terms from the product: $$-\frac{1}{3} x^{3} + \frac{1}{4} x^{2} + \frac{1}{2} x^{2} - \frac{3}{8} x - \frac{2}{3} x + \frac{1}{2}$$ 4. Combine like terms: For $x^{3}$: $-\frac{1}{3} x^{3}$ For $x^{2}$: $\frac{1}{4} x^{2} + \frac{1}{2} x^{2} = \frac{1}{4} x^{2} + \frac{2}{4} x^{2} = \frac{3}{4} x^{2}$ For $x$: $-\frac{3}{8} x - \frac{2}{3} x = -\frac{9}{24} x - \frac{16}{24} x = -\frac{25}{24} x$ Constant: $\frac{1}{2}$ 5. Now subtract the second parentheses: $$-\left(\frac{1}{3} x^{2} - x\right) = -\frac{1}{3} x^{2} + x$$ 6. Add this to the previous result: $$-\frac{1}{3} x^{3} + \frac{3}{4} x^{2} - \frac{25}{24} x + \frac{1}{2} - \frac{1}{3} x^{2} + x$$ 7. Combine like terms again: For $x^{3}$: $-\frac{1}{3} x^{3}$ For $x^{2}$: $\frac{3}{4} x^{2} - \frac{1}{3} x^{2} = \frac{9}{12} x^{2} - \frac{4}{12} x^{2} = \frac{5}{12} x^{2}$ For $x$: $-\frac{25}{24} x + x = -\frac{25}{24} x + \frac{24}{24} x = -\frac{1}{24} x$ Constant: $\frac{1}{2}$ 8. Final simplified expression for (b): $$-\frac{1}{3} x^{3} + \frac{5}{12} x^{2} - \frac{1}{24} x + \frac{1}{2}$$