If `(x+a)` is a factor of both the quadratic polynomials `x^(2) + px+ q and x^(2) + lx + m` , where p,q,l and m are constants, then which one of the following is correct?

Options

A. ` a= (m-q) //(l-p) (l!=p)`
B. ` a = (m-q)// (l+p) (l !=-p)`
C. ` l = (m-q)//(a-p) ( a != p)`
D. ` p= ( m - q) // (a-l) ( a !=l)`

Explanation

To solve the problem, we need to determine the relationship between the constants \( p, q, l, \) and \( m \) given that \( (x + a) \) is a factor of both quadratic polynomials \( x^2 + px + q \) and \( x^2 + lx + m \).

### Step-by-Step Solution:

1. **Identify the Roots**: Since \( (x + a) \) is a factor of both polynomials, it implies that \( x = -a \) is a root of both polynomials.

2. **Substituting in the First Polynomial**: Substitute \( x = -a \) into the first polynomial:
\[ (-a)^2 + p(-a) + q = 0 \]
Simplifying this gives:
\[ a^2 - ap + q = 0 \quad \text{(Equation 1)} \]

3. **Substituting in the Second Polynomial**: Now substitute \( x = -a \) into the second polynomial:
\[ (-a)^2 + l(-a) + m = 0 \]
Simplifying this gives:
\[ a^2 - al + m = 0 \quad \text{(Equation 2)} \]

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

5. **Eliminating \( a^2 \)**: Subtract \( a^2 \) from both sides:
\[ -ap + q = -al + m \]

6. **Rearranging the Equation**: Rearranging gives:
\[ al - ap + q - m = 0 \]
This can be rewritten as:
\[ a(l - p) = m - q \quad \text{(Equation 3)} \]

7. **Solving for \( a \)**: From Equation 3, we can express \( a \) as:
\[ a = \frac{m - q}{l - p} \quad \text{(provided \( l \neq p \))} \]

### Conclusion:
Thus, the correct relationship is:
\[ a = \frac{m - q}{l - p} \quad \text{where } l \neq p \]