If `x_1, x_2, x_3,........x_n` are the roots of `x^n + ax + b = 0`, then the value of `(x_1 - x_2)(x_1 - x_3) (x_1 - x_4) .......(x_1 – x_n)` is equal to

Options

A. `nx _(1) +b`
B. ` n x _(1)^(n-1) +a`
C. `nx _(1) ^(n-1)`
D. `nx _(1) ^(n-1)`

Answer

To solve the problem, we need to find the value of the expression \((x_1 - x_2)(x_1 - x_3)(x_1 - x_4) \cdots (x_1 - x_n)\) where \(x_1, x_2, x_3, \ldots, x_n\) are the roots of the polynomial \(x^n + ax + b = 0\).

### Step-by-Step Solution:

1. **Understanding the Polynomial**:
The polynomial can be expressed in factored form as:
\[ x^n + ax + b = (x - x_1)(x - x_2)(x - x_3) \cdots (x - x_n) \]
This means that \(x_1, x_2, \ldots, x_n\) are the roots of the polynomial.

2. **Setting Up the Expression**:
We want to evaluate:
\[ (x_1 - x_2)(x_1 - x_3)(x_1 - x_4) \cdots (x_1 - x_n) \]
This expression can be interpreted as the product of the differences between \(x_1\) and the other roots.

3. **Using Limits**:
To find this product, we can use the fact that:
\[ (x - x_1)(x - x_2)(x - x_3) \cdots (x - x_n) = x^n + ax + b \]
We can take the limit as \(x\) approaches \(x_1\):
\[ \lim_{x \to x_1} \frac{x^n + ax + b}{x - x_1} \]

4. **Applying L'Hôpital's Rule**:
Since substituting \(x = x_1\) gives us the indeterminate form \(0/0\), we can apply L'Hôpital's Rule:
\[ \lim_{x \to x_1} \frac{x^n + ax + b}{x - x_1} = \lim_{x \to x_1} \frac{n x^{n-1} + a}{1} \]

5. **Calculating the Limit**:
Now we substitute \(x = x_1\):
\[ n x_1^{n-1} + a \]

6. **Final Result**:
Therefore, the value of the expression \((x_1 - x_2)(x_1 - x_3)(x_1 - x_4) \cdots (x_1 - x_n)\) is:
\[ n x_1^{n-1} + a \]