Expand `(3x^2- x/2)^5`

Answer

To expand the expression \((3x^2 - \frac{x}{2})^5\), we will use the Binomial Theorem, which states that:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

In our case, \(a = 3x^2\), \(b = -\frac{x}{2}\), and \(n = 5\).

### Step-by-Step Solution:

1. **Identify \(a\), \(b\), and \(n\)**:
- \(a = 3x^2\)
- \(b = -\frac{x}{2}\)
- \(n = 5\)

2. **Apply the Binomial Theorem**:
\[ (3x^2 - \frac{x}{2})^5 = \sum_{k=0}^{5} \binom{5}{k} (3x^2)^{5-k} \left(-\frac{x}{2}\right)^k \]

3. **Calculate each term in the expansion**:
- For \(k = 0\):
\[ \binom{5}{0} (3x^2)^5 \left(-\frac{x}{2}\right)^0 = 1 \cdot (3x^2)^5 \cdot 1 = 243x^{10} \]
- For \(k = 1\):
\[ \binom{5}{1} (3x^2)^4 \left(-\frac{x}{2}\right)^1 = 5 \cdot (3^4)(x^8) \cdot \left(-\frac{x}{2}\right) = -5 \cdot 81x^8 \cdot \frac{x}{2} = -\frac{405}{2} x^9 \]
- For \(k = 2\):
\[ \binom{5}{2} (3x^2)^3 \left(-\frac{x}{2}\right)^2 = 10 \cdot (3^3)(x^6) \cdot \left(\frac{x^2}{4}\right) = 10 \cdot 27x^6 \cdot \frac{x^2}{4} = \frac{270}{4} x^8 = \frac{135}{2} x^8 \]
- For \(k = 3\):
\[ \binom{5}{3} (3x^2)^2 \left(-\frac{x}{2}\right)^3 = 10 \cdot (3^2)(x^4) \cdot \left(-\frac{x^3}{8}\right) = -10 \cdot 9x^4 \cdot \frac{x^3}{8} = -\frac{90}{8} x^7 = -\frac{45}{4} x^7 \]
- For \(k = 4\):
\[ \binom{5}{4} (3x^2)^1 \left(-\frac{x}{2}\right)^4 = 5 \cdot (3)(x^2) \cdot \left(\frac{x^4}{16}\right) = 5 \cdot 3x^2 \cdot \frac{x^4}{16} = \frac{15}{16} x^6 \]
- For \(k = 5\):
\[ \binom{5}{5} (3x^2)^0 \left(-\frac{x}{2}\right)^5 = 1 \cdot 1 \cdot \left(-\frac{x^5}{32}\right) = -\frac{x^5}{32} \]

4. **Combine all the terms**:
\[ (3x^2 - \frac{x}{2})^5 = 243x^{10} - \frac{405}{2} x^9 + \frac{135}{2} x^8 - \frac{45}{4} x^7 + \frac{15}{16} x^6 - \frac{x^5}{32} \]

### Final Expanded Form:
\[ (3x^2 - \frac{x}{2})^5 = 243x^{10} - \frac{405}{2} x^9 + \frac{135}{2} x^8 - \frac{45}{4} x^7 + \frac{15}{16} x^6 - \frac{x^5}{32} \]