1. **Problem: Expand and simplify $ (1 - x + x^2)^3 $.** 2. To expand this cube, we use the binomial expansion for a trinomial power. We write: $$ (1 - x + x^2)^3 = (1 - x + x^2)(1 - x + x^2)(1 - x + x^2) $$ 3. First, find the square of the trinomial: $$ (1 - x + x^2)^2 = (1 - x + x^2)(1 - x + x^2) $$ 4. Multiply term by term: - $1 \cdot 1 = 1$ - $1 \cdot (-x) = -x$ - $1 \cdot x^2 = x^2$ - $(-x) \cdot 1 = -x$ - $(-x) \cdot (-x) = x^2$ - $(-x) \cdot x^2 = -x^3$ - $x^2 \cdot 1 = x^2$ - $x^2 \cdot (-x) = -x^3$ - $x^2 \cdot x^2 = x^4$ 5. Combine like terms: $$1 - x - x + x^2 + x^2 + x^2 - x^3 - x^3 + x^4 = 1 - 2x + 3x^2 - 2x^3 + x^4$$ 6. Now multiply this result by $(1 - x + x^2)$ again: $$ (1 - 2x + 3x^2 - 2x^3 + x^4)(1 - x + x^2) $$ 7. Multiply each term in the first polynomial by each term in the second: - $1 \cdot 1 = 1$ - $1 \cdot (-x) = -x$ - $1 \cdot x^{2} = x^{2}$ - $-2x \cdot 1 = -2x$ - $-2x \cdot (-x) = 2x^{2}$ - $-2x \cdot x^{2} = -2x^{3}$ - $3x^{2} \cdot 1 = 3x^{2}$ - $3x^{2} \cdot (-x) = -3x^{3}$ - $3x^{2} \cdot x^{2} = 3x^{4}$ - $-2x^{3} \cdot 1 = -2x^{3}$ - $-2x^{3} \cdot (-x) = 2x^{4}$ - $-2x^{3} \cdot x^{2} = -2x^{5}$ - $x^{4} \cdot 1 = x^{4}$ - $x^{4} \cdot (-x) = -x^{5}$ - $x^{4} \cdot x^{2} = x^{6}$ 8. Now combine like terms: $$ 1 - x - 2x + x^{2} + 2x^{2} + 3x^{2} - 2x^{3} - 3x^{3} - 2x^{3} + 3x^{4} + 2x^{4} + x^{4} - 2x^{5} - x^{5} + x^{6} $$ 9. Group and sum coefficients: - Constant: $1$ - $x$: $-x - 2x = -3x$ - $x^{2}$: $x^{2} + 2x^{2} + 3x^{2} = 6x^{2}$ - $x^{3}$: $-2x^{3} - 3x^{3} - 2x^{3} = -7x^{3}$ - $x^{4}$: $3x^{4} + 2x^{4} + x^{4} = 6x^{4}$ - $x^{5}$: $-2x^{5} - x^{5} = -3x^{5}$ - $x^{6}$: $x^{6}$ 10. Final expanded form: $$ (1 - x + x^{2})^{3} = 1 - 3x + 6x^{2} - 7x^{3} + 6x^{4} - 3x^{5} + x^{6} $$ --- 11. **Problem: Evaluate $f(x) = 3(2x + 1)$ at $x=4$.** 12. Plug in $x=4$: $$ f(4) = 3(2(4) + 1) = 3(8 + 1) = 3(9) = 27 $$ --- 13. **Problem: Find $f(k+2)$ for $f(x) = x^{2} + 3x + 5$.** 14. Substitute $x = k + 2$: $$ f(k+2) = (k+2)^{2} + 3(k+2) + 5 = (k^{2} + 4k + 4) + 3k + 6 + 5 $$ 15. Combine like terms: $$ k^{2} + 4k + 4 + 3k + 6 + 5 = k^{2} + 7k + 15 $$ --- 16. **Problem: Add $f(x) = 3x^{2} - 4x +8$ and $g(x) = 5x + 6$.** 17. Add corresponding terms: $$ (3x^{2} - 4x + 8) + (5x + 6) = 3x^{2} + (-4x + 5x) + (8 + 6) = 3x^{2} + x + 14 $$ --- 18. **Problem: Subtract $f(x) = 3x^{2} - 6x - 4$ from $g(x) = -2x^{2} + x + 5$.** 19. Compute $g(x) - f(x)$: $$ (-2x^{2} + x + 5) - (3x^{2} - 6x - 4) = -2x^{2} + x + 5 - 3x^{2} + 6x + 4 $$ 20. Combine like terms: $$ (-2x^{2} - 3x^{2}) + (x + 6x) + (5 + 4) = -5x^{2} + 7x + 9 $$ --- 21. **Problem: Find the product of $f(x) = x - 3$ and $g(x) = 2x - 9$.** 22. Multiply: $$ (x - 3)(2x - 9) = x imes 2x - x imes 9 - 3 imes 2x + (-3) imes (-9) = 2x^{2} - 9x - 6x + 27 $$ 23. Combine like terms: $$ 2x^{2} - 15x + 27 $$ --- 24. **Problem: Divide $f(x) = 35x^{6} - 10x^{4} + 5x^{3}$ by $g(x) = 5x^{3}$.** 25. Divide each term separately: $$ \frac{35x^{6}}{5x^{3}} = 7x^{3}, \quad \frac{-10x^{4}}{5x^{3}} = -2x, \quad \frac{5x^{3}}{5x^{3}} = 1 $$ 26. So, $$ \frac{f(x)}{g(x)} = 7x^{3} - 2x + 1 $$ --- 27. **Problem: Given $g(x) = 2x - 1$ and $f(x) = x^{2} + 6$, find $(g \circ f)(x) = g(f(x))$.** 28. Substitute $f(x)$ into $g$: $$ g(f(x)) = 2(f(x)) - 1 = 2(x^{2} + 6) - 1 = 2x^{2} + 12 - 1 = 2x^{2} + 11 $$ --- 29. **Problem: Given $f(x) = 2x + 3$, find its inverse $f^{-1}(x)$ and verify.** 30. To find $f^{-1}(x)$, replace $f(x)$ with $y$: $$ y = 2x + 3 $$ 31. Swap $x$ and $y$ to solve for $y$: $$ x = 2y + 3 $$ 32. Solve for $y$: $$ 2y = x - 3 \Rightarrow y = \frac{x - 3}{2} $$ 33. So, inverse function: $$ f^{-1}(x) = \frac{x - 3}{2} $$ 34. Verify by composition: $$ f(f^{-1}(x)) = 2 \times \frac{x - 3}{2} + 3 = (x - 3) + 3 = x $$ $$ f^{-1}(f(x)) = \frac{(2x + 3) - 3}{2} = \frac{2x}{2} = x $$ 35. Both confirm correctness. --- 36. **Problem: If $3^{x} = 9$, solve for $x$.** 37. Note $9 = 3^{2}$, so $$ 3^{x} = 3^{2} \Rightarrow x = 2 $$ --- 38. **Problem: If $32^{x} = 0.0625$, solve for $x$.** 39. $32 = 2^{5}$. Also, $$ 0.0625 = \frac{1}{16} = 2^{-4} $$ 40. Rewrite equation: $$ (2^{5})^{x} = 2^{-4} \Rightarrow 2^{5x} = 2^{-4} \Rightarrow 5x = -4 \Rightarrow x = -\frac{4}{5} $$ --- 41. **Problem: If $4^{6(x - 3)} = 1$, solve for $x$.** 42. Since $4^{0} = 1$, the exponent must be zero: $$ 6(x - 3) = 0 \Rightarrow x - 3 = 0 \Rightarrow x = 3 $$