If `(x+a)` is the factor of the polynomials `(x^(2)+px+q)` and `(x^(2)+mx+n)` , then the value of 'a' is

Options

A. `(n-q)/(m-p)`
B. `(m-p)/(n-q)`
C. `(q-n)/(m-p)`
D. `(m-p)/(q-n)`

Answer

To find the value of 'a' given that `(x + a)` is a factor of the polynomials `(x^2 + px + q)` and `(x^2 + mx + n)`, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Factor Condition**:
Since `(x + a)` is a factor of both polynomials, substituting `x = -a` into each polynomial should yield a remainder of zero.

2. **Substituting in the First Polynomial**:
For the polynomial `(x^2 + px + q)`, substituting `x = -a` gives:
\[ (-a)^2 + p(-a) + q = 0 \]
Simplifying this, we get:
\[ a^2 - pa + q = 0 \quad \text{(Equation 1)} \]

3. **Substituting in the Second Polynomial**:
For the polynomial `(x^2 + mx + n)`, substituting `x = -a` gives:
\[ (-a)^2 + m(-a) + n = 0 \]
Simplifying this, we get:
\[ a^2 - ma + n = 0 \quad \text{(Equation 2)} \]

4. **Setting the Two Equations Equal**:
Since both equations equal zero, we can set them equal to each other:
\[ a^2 - pa + q = a^2 - ma + n \]

5. **Cancelling \(a^2\)**:
We can cancel \(a^2\) from both sides:
\[ -pa + q = -ma + n \]

6. **Rearranging the Equation**:
Rearranging gives us:
\[ -pa + ma = n - q \]
Factoring out \(a\) from the left side:
\[ a(m - p) = n - q \]

7. **Solving for 'a'**:
Finally, we can solve for \(a\):
\[ a = \frac{n - q}{m - p} \]

### Final Answer:
Thus, the value of 'a' is:
\[ a = \frac{n - q}{m - p} \]