2) \( (x-2)^{4} \)

To expand \( (x - 2)^4 \), you can use the **Binomial Theorem**, which states: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] For \( (x - 2)^4 \): - \( a = x \) - \( b = -2 \) - \( n = 4 \) Let's expand it step by step: \[ (x - 2)^4 = \sum_{k=0}^{4} \binom{4}{k} x^{4-k} (-2)^k \] Calculate each term: 1. **When \( k = 0 \):** \[ \binom{4}{0} x^4 (-2)^0 = 1 \cdot x^4 \cdot 1 = x^4 \] 2. **When \( k = 1 \):** \[ \binom{4}{1} x^3 (-2)^1 = 4 \cdot x^3 \cdot (-2) = -8x^3 \] 3. **When \( k = 2 \):** \[ \binom{4}{2} x^2 (-2)^2 = 6 \cdot x^2 \cdot 4 = 24x^2 \] 4. **When \( k = 3 \):** \[ \binom{4}{3} x^1 (-2)^3 = 4 \cdot x \cdot (-8) = -32x \] 5. **When \( k = 4 \):** \[ \binom{4}{4} x^0 (-2)^4 = 1 \cdot 1 \cdot 16 = 16 \] **Combine all the terms:** \[ (x - 2)^4 = x^4 - 8x^3 + 24x^2 - 32x + 16 \] **Final Expanded Form:** \[ x^4 - 8x^3 + 24x^2 - 32x + 16 \]