Find the coefficients of `x^(4)` in `(1 - x)^(2) (2 + x)^(5)` using binomial theorem.

Answer

To find the coefficient of \(x^4\) in the expression \((1 - x)^2 (2 + x)^5\) using the Binomial Theorem, we will break down the problem step by step.

### Step 1: Expand \((1 - x)^2\)
Using the Binomial Theorem:
\[ (1 - x)^2 = \sum_{k=0}^{2} \binom{2}{k} (1)^{2-k} (-x)^k \]
Calculating the terms:
- For \(k=0\): \(\binom{2}{0} (1)^2 (-x)^0 = 1\)
- For \(k=1\): \(\binom{2}{1} (1)^1 (-x)^1 = -2x\)
- For \(k=2\): \(\binom{2}{2} (1)^0 (-x)^2 = x^2\)

Thus,
\[ (1 - x)^2 = 1 - 2x + x^2 \]

### Step 2: Expand \((2 + x)^5\)
Using the Binomial Theorem:
\[ (2 + x)^5 = \sum_{j=0}^{5} \binom{5}{j} (2)^{5-j} (x)^j \]
Calculating the relevant terms:
- For \(j=0\): \(\binom{5}{0} (2)^5 (x)^0 = 32\)
- For \(j=1\): \(\binom{5}{1} (2)^4 (x)^1 = 80x\)
- For \(j=2\): \(\binom{5}{2} (2)^3 (x)^2 = 80x^2\)
- For \(j=3\): \(\binom{5}{3} (2)^2 (x)^3 = 40x^3\)
- For \(j=4\): \(\binom{5}{4} (2)^1 (x)^4 = 10x^4\)
- For \(j=5\): \(\binom{5}{5} (2)^0 (x)^5 = x^5\)

Thus,
\[ (2 + x)^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5 \]

### Step 3: Multiply the two expansions
Now we need to find the coefficient of \(x^4\) in the product \((1 - 2x + x^2)(32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5)\).

We will find the contributions to \(x^4\) from each term:
1. From \(1\) in \((1 - 2x + x^2)\) and \(10x^4\) in \((2 + x)^5\):
- Contribution: \(1 \cdot 10 = 10\)

2. From \(-2x\) in \((1 - 2x + x^2)\) and \(40x^3\) in \((2 + x)^5\):
- Contribution: \(-2 \cdot 40 = -80\)

3. From \(x^2\) in \((1 - 2x + x^2)\) and \(80x^2\) in \((2 + x)^5\):
- Contribution: \(1 \cdot 80 = 80\)

Now, we sum these contributions:
\[ 10 - 80 + 80 = 10 \]

### Final Result
The coefficient of \(x^4\) in \((1 - x)^2 (2 + x)^5\) is **10**.

---