If (x+k) is the HCF of `(x^(2) + ax + b)` and `(x^(2) + px + q)`, then the value of k is

Options

A. `(a-b)/(p+q)`
B. `(q-b)/(p-a)`
C. `(b-q)/(p-a)`
D. `(q-b)/(a-p)`

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Problem
We are given that \( (x + k) \) is the HCF of the polynomials \( (x^2 + ax + b) \) and \( (x^2 + px + q) \). This means that both polynomials must equal zero when \( x = -k \).

### Step 2: Substitute \( x = -k \)
Substituting \( x = -k \) into both polynomials gives us:
1. For the first polynomial:
\[ (-k)^2 + a(-k) + b = 0 \]
Simplifying this, we get:
\[ k^2 - ak + b = 0 \quad \text{(1)} \]

2. For the second polynomial:
\[ (-k)^2 + p(-k) + q = 0 \]
Simplifying this, we get:
\[ k^2 - pk + q = 0 \quad \text{(2)} \]

### Step 3: Set Up the Equations
From equations (1) and (2), we have:
1. \( k^2 - ak + b = 0 \)
2. \( k^2 - pk + q = 0 \)

### Step 4: Subtract the Two Equations
Now, we can subtract equation (1) from equation (2):
\[ (k^2 - pk + q) - (k^2 - ak + b) = 0 \]
This simplifies to:
\[ -ak + b - (-pk + q) = 0 \]
Rearranging gives:
\[ pk - ak = q - b \]
Factoring out \( k \) from the left side:
\[ k(p - a) = q - b \]

### Step 5: Solve for \( k \)
Now, we can solve for \( k \):
\[ k = \frac{q - b}{p - a} \]

### Final Answer
Thus, the value of \( k \) is:
\[ k = \frac{q - b}{p - a} \]

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