If HCF of `(x^(2)+x-12) and (2x^(2)-kx -9)` is `(x-k)`, then value of k is

Options

A. `-3 `
B. 3
C. `-4`
D. 4

Answer

To solve the problem, we need to find the value of \( k \) given that the HCF of the polynomials \( P(x) = x^2 + x - 12 \) and \( Q(x) = 2x^2 - kx - 9 \) is \( x - k \).

### Step 1: Factor the polynomial \( P(x) \)

First, we will factor \( P(x) = x^2 + x - 12 \).

To factor, we look for two numbers that multiply to \(-12\) (the constant term) and add to \(1\) (the coefficient of \(x\)). The numbers \(4\) and \(-3\) satisfy these conditions.

Thus, we can write:
\[ P(x) = (x + 4)(x - 3) \]

### Step 2: Set up the condition for HCF

Since the HCF of \( P(x) \) and \( Q(x) \) is given as \( x - k \), this means that \( x - k \) must be a factor of both \( P(x) \) and \( Q(x) \).

From the factorization of \( P(x) \), the possible roots (which are the values of \( x \) that make \( P(x) = 0 \)) are:
\[ x = -4 \quad \text{and} \quad x = 3 \]

This implies that \( k \) can be either \( -4 \) or \( 3 \).

### Step 3: Substitute \( k \) into \( Q(x) \)

Now, we will check each case to find the correct value of \( k \).

1. **Case 1: \( k = -4 \)**

Substituting \( k = -4 \) into \( Q(x) \):
\[ Q(x) = 2x^2 - (-4)x - 9 = 2x^2 + 4x - 9 \]

Now, we need to check if \( x + 4 \) is a factor of \( Q(x) \). We can use synthetic division or direct substitution to check if \( Q(-4) = 0 \):
\[ Q(-4) = 2(-4)^2 + 4(-4) - 9 = 32 - 16 - 9 = 7 \quad (\text{not a factor}) \]

2. **Case 2: \( k = 3 \)**

Substituting \( k = 3 \) into \( Q(x) \):
\[ Q(x) = 2x^2 - 3x - 9 \]

Now, we need to check if \( x - 3 \) is a factor of \( Q(x) \). We can use synthetic division or direct substitution to check if \( Q(3) = 0 \):
\[ Q(3) = 2(3)^2 - 3(3) - 9 = 18 - 9 - 9 = 0 \quad (\text{is a factor}) \]

### Conclusion

Since \( k = 3 \) satisfies the condition that \( x - k \) is a factor of both \( P(x) \) and \( Q(x) \), we conclude that the value of \( k \) is:
\[ \boxed{3} \]